Put Marbles in Bags
10 Prerequisite LeetCode Problems
“Put Marbles in Bags” involves dynamic programming, array manipulation and understanding of mathematical properties of numbers. Here are 10 simpler problems to build these foundational skills:
“Best Time to Buy and Sell Stock” (LeetCode Problem #121): This problem practices array manipulation and basic dynamic programming.
“Partition Equal Subset Sum” (LeetCode Problem #416): This problem practices dynamic programming concepts and array manipulation.
“Minimum Size Subarray Sum” (LeetCode Problem #209): This problem deals with array manipulations and introduces the sliding window concept which might be beneficial for the problem at hand.
“Maximum Subarray” (LeetCode Problem #53): This problem introduces the concept of dynamic programming on arrays.
“Climbing Stairs” (LeetCode Problem #70): This problem introduces the basics of dynamic programming.
“Coin Change” (LeetCode Problem #322): This problem introduces the concept of dynamic programming with state transition.
“Minimum Path Sum” (LeetCode Problem #64): This problem provides practice in dynamic programming while working on a 2D grid, which is a useful concept for “Put Marbles in Bags”.
“Unique Paths” (LeetCode Problem #62): This problem is a good introduction to the concept of navigating a grid, similar to the distribution of marbles into bags.
“Candy” (LeetCode Problem #135): This problem involves distributing items (candies) based on certain conditions, which might be helpful for understanding the distribution of marbles in bags.
“Split Array Largest Sum” (LeetCode Problem #410): This problem involves splitting an array into certain parts with the aim of minimizing the maximum sum, which is similar to “Put Marbles in Bags”.
Brush up on dynamic programming strategies, specifically for array partitioning problems.
Problem Classification
Problem Statement: You have k bags. You are given a 0indexed integer array weights where weights[i] is the weight of the ith marble. You are also given the integer k.
Divide the marbles into the k bags according to the following rules:
No bag is empty. If the ith marble and jth marble are in a bag, then all marbles with an index between the ith and jth indices should also be in that same bag. If a bag consists of all the marbles with an index from i to j inclusively, then the cost of the bag is weights[i] + weights[j]. The score after distributing the marbles is the sum of the costs of all the k bags.
Return the difference between the maximum and minimum scores among marble distributions.
Example 1:
Input: weights = [1,3,5,1], k = 2 Output: 4 Explanation: The distribution [1],[3,5,1] results in the minimal score of (1+1) + (3+1) = 6. The distribution [1,3],[5,1], results in the maximal score of (1+3) + (5+1) = 10. Thus, we return their difference 10  6 = 4. Example 2:
Input: weights = [1, 3], k = 2 Output: 0 Explanation: The only distribution possible is [1],[3]. Since both the maximal and minimal score are the same, we return 0.
Constraints:
1 <= k <= weights.length <= 105 1 <= weights[i] <= 109
Analyze the provided problem statement. Categorize it based on its domain, ignoring ‘How’ it might be solved. Identify and list out the ‘What’ components. Based on these, further classify the problem. Explain your categorizations.
Visual Model of the Problem
How to visualize the problem statement for this problem?
Problem Restatement
Could you start by paraphrasing the problem statement in your own words? Try to distill the problem into its essential elements and make sure to clarify the requirements and constraints. This exercise should aid in understanding the problem better and aligning our thought process before jumping into solving it.
Abstract Representation of the Problem
Could you help me formulate an abstract representation of this problem?
Alternatively, if you’re working on a specific problem, you might ask something like:
Given this problem, how can we describe it in an abstract way that emphasizes the structure and key elements, without the specific realworld details?
Terminology
Are there any specialized terms, jargon, or technical concepts that are crucial to understanding this problem or solution? Could you define them and explain their role within the context of this problem?
Problem Simplification and Explanation
Could you please break down this problem into simpler terms? What are the key concepts involved and how do they interact? Can you also provide a metaphor or analogy to help me understand the problem better?
Constraints
Given the problem statement and the constraints provided, identify specific characteristics or conditions that can be exploited to our advantage in finding an efficient solution. Look for patterns or specific numerical ranges that could be useful in manipulating or interpreting the data.
What are the key insights from analyzing the constraints?
Case Analysis
Could you please provide additional examples or test cases that cover a wider range of the input space, including edge and boundary conditions? In doing so, could you also analyze each example to highlight different aspects of the problem, key constraints and potential pitfalls, as well as the reasoning behind the expected output for each case? This should help in generating key insights about the problem and ensuring the solution is robust and handles all possible scenarios.
Identification of Applicable Theoretical Concepts
Can you identify any mathematical or algorithmic concepts or properties that can be applied to simplify the problem or make it more manageable? Think about the nature of the operations or manipulations required by the problem statement. Are there existing theories, metrics, or methodologies in mathematics, computer science, or related fields that can be applied to calculate, measure, or perform these operations more effectively or efficiently?
Problem Breakdown and Solution Methodology
Given the problem statement, can you explain in detail how you would approach solving it? Please break down the process into smaller steps, illustrating how each step contributes to the overall solution. If applicable, consider using metaphors, analogies, or visual representations to make your explanation more intuitive. After explaining the process, can you also discuss how specific operations or changes in the problem’s parameters would affect the solution? Lastly, demonstrate the workings of your approach using one or more example cases.
Inference of ProblemSolving Approach from the Problem Statement
How did you infer from the problem statement that this problem can be solved using ?
Stepwise Refinement
Could you please provide a stepwise refinement of our approach to solving this problem?
How can we take the highlevel solution approach and distill it into more granular, actionable steps?
Could you identify any parts of the problem that can be solved independently?
Are there any repeatable patterns within our solution?
Solution Approach and Analysis
Given the problem statement, can you explain in detail how you would approach solving it? Please break down the process into smaller steps, illustrating how each step contributes to the overall solution. If applicable, consider using metaphors, analogies, or visual representations to make your explanation more intuitive. After explaining the process, can you also discuss how specific operations or changes in the problem’s parameters would affect the solution? Lastly, demonstrate the workings of your approach using one or more example cases.
Thought Process
Explain the thought process by thinking step by step to solve this problem from the problem statement and code the final solution. Write code in Python3. What are the cues in the problem statement? What direction does it suggest in the approach to the problem? Generate insights about the problem statement.
From Brute Force to Optimal Solution
Could you please begin by illustrating a brute force solution for this problem? After detailing and discussing the inefficiencies of the brute force approach, could you then guide us through the process of optimizing this solution? Please explain each step towards optimization, discussing the reasoning behind each decision made, and how it improves upon the previous solution. Also, could you show how these optimizations impact the time and space complexity of our solution?
Coding Constructs
Consider the following piece of complex software code.
What are the highlevel problemsolving strategies or techniques being used by this code?
If you had to explain the purpose of this code to a nonprogrammer, what would you say?
Can you identify the logical elements or constructs used in this code, independent of any programming language?
Could you describe the algorithmic approach used by this code in plain English?
What are the key steps or operations this code is performing on the input data, and why?
Can you identify the algorithmic patterns or strategies used by this code, irrespective of the specific programming language syntax?
Language Agnostic Coding Drills
Your mission is to deconstruct this code into the smallest possible learning units, each corresponding to a separate coding concept. Consider these concepts as unique coding drills that can be individually implemented and later assembled into the final solution.
Dissect the code and identify each distinct concept it contains. Remember, this process should be languageagnostic and generally applicable to most modern programming languages.
Once you’ve identified these coding concepts or drills, list them out in order of increasing difficulty. Provide a brief description of each concept and why it is classified at its particular difficulty level.
Next, describe the problemsolving approach that would lead from the problem statement to the final solution. Think about how each of these coding drills contributes to the overall solution. Elucidate the stepbystep process involved in using these drills to solve the problem. Please refrain from writing any actual code; we’re focusing on understanding the process and strategy.
Targeted Drills in Python
Now that you’ve identified and ordered the coding concepts from a complex software code in the previous exercise, let’s focus on creating Pythonbased coding drills for each of those concepts.
Begin by writing a separate piece of Python code that encapsulates each identified concept. These individual drills should illustrate how to implement each concept in Python. Please ensure that these are suitable even for those with a basic understanding of Python.
In addition to the general concepts, identify and write coding drills for any problemspecific concepts that might be needed to create a solution. Describe why these drills are essential for our problem.
Once all drills have been coded, describe how these pieces can be integrated together in the right order to solve the initial problem. Each drill should contribute to building up to the final solution.
Remember, the goal is to not only to write these drills but also to ensure that they can be cohesively assembled into one comprehensive solution.
Q&A
Similar Problems
Given the problem [provide the problem], identify and list down 10 similar problems on LeetCode. These should cover similar concepts or require similar problemsolving approaches as the provided problem. Please also give a brief reason as to why you think each problem is similar to the given problem.
10 Prerequisite LeetCode Problems
This problem involves elements of dynamic programming and range queries, which can often be found in Segment Trees or Binary Indexed Tree problems. Also, it involves the principle of choosing partitions which require understanding of sorting and greedy algorithms.
Here are some fundamental problems that should help you understand these concepts better:
Dynamic Programming
 Climbing Stairs (LeetCode #70)
 House Robber (LeetCode #198)
 Best Time to Buy and Sell Stock (LeetCode #121)
Range Queries
 Range Sum Query  Immutable (LeetCode #303)
 Range Sum Query 2D  Immutable (LeetCode #304)
Segment Trees / Binary Indexed Trees
 Range Sum Query  Mutable (LeetCode #307)
 Maximum of Absolute Value Expression (LeetCode #1131)
Sorting and Greedy Algorithms
 Assign Cookies (LeetCode #455)
 Nonoverlapping Intervals (LeetCode #435)
 Task Scheduler (LeetCode #621)
For more practice, consider problems on partitions and divideandconquer. These problems can be complex, but they often provide insight into how to approach problems like “Put Marbles in Bags”.
Please note that these problems may not directly map to the “Put Marbles in Bags” problem. However, understanding and being able to solve these problems will likely give you a better understanding of the key strategies and data structures that can be used to solve the given problem.
Problem Analysis and Key Insights
Analyzing the problem statement gives us key insights into the problem’s nature and what we need to do to solve it. Here are some key insights:
Array Partitioning:
The problem involves dividing the weights array into ‘k’ partitions, each partition representing a bag. The nature of the partitions (order matters, no bag is empty) suggest this is a partitioning problem with specific conditions.
Bounded Cost Calculation:
The cost of each bag is defined as the sum of the weight of the first and the last marble in the bag. This is a specific rule that we need to take into account when calculating the cost. It tells us that we only need to consider the first and last elements of each partition.
Minimum and Maximum Scores:
We are asked to find the difference between the maximum and minimum scores, which indicates that for each possible distribution of marbles, we need to compute two separate scores (maximum and minimum).
Order Matters:
The rule that if two marbles are in the same bag, all marbles between them must also be in that bag, indicates that the order of marbles matters in this problem. This suggests that algorithms that preserve the original order of elements (like dynamic programming) might be helpful.
Difference in Scores:
Ultimately, we need to return the difference between the maximum and minimum total scores. This indicates that we must keep track of these two values throughout the computation process to eventually find their difference.
Constraints:
The constraints on the size of the weights array and the value of the weights indicate that the solution must be efficient enough to handle large inputs.
These insights will be useful in identifying appropriate strategies and algorithms for solving the problem.
Problem Boundary
The scope of the problem “Put Marbles in Bags” can be defined by the range of the following parameters:
Marbles (weights): An array of integers where
weights[i]
is the weight of theith
marble. The constraints mention that the array length is between 1 and 105, with each weight between 1 and 109.Bags (k): An integer which represents the number of bags to be filled with marbles. The constraints indicate that
k
is between 1 and the length of theweights
array.
The problem requires you to partition the weights
array into k
contiguous subarrays (representing bags) such that no bag is empty and the cost of each bag is defined as the sum of the weight of the first marble and the weight of the last marble in that bag. The goal is to find the difference between the maximum and minimum possible scores, where the score is the sum of costs of all k
bags.
To solve this problem, you need to understand how to:
 Apply dynamic programming concepts, particularly in relation to array partitioning problems.
 Use range queries for sum calculation within certain index ranges.
 Efficiently track minimum/maximum values within a sliding window, which could be implemented using data structures like deques or segment trees.
 Implement a solution within the provided constraints to ensure the program runs efficiently.
The skills developed while solving this problem will also be helpful in dealing with other similar problems related to array manipulations, partitioning, and dynamic programming.
The problem can be viewed as a dynamic programming problem. You can use the following way to establish boundaries:
State: Define
dp[i][j]
as the minimum cost to partition the firsti
marbles intoj
bags. The range ofi
is [1, N] (N is the size of the weights array) and the range ofj
is [1, k].Base case: For
dp[i][1]
(only 1 bag), the cost is just the weight of the first marble plus the weight of the last marble, i.e.,weights[0] + weights[i1]
fori
in [1, N].Transition: For
j
> 1, we need to find the optimal partition pointp
for the last bag that minimizes the cost. So for eachi
>j
,dp[i][j] = min(dp[p][j1] + weights[p] + weights[i1])
forp
in [j1, i1]. This tries every possible partition point for the last bag and picks the one with the minimum cost.Final answer: The minimum score is
dp[N][k]
, and the maximum score could be calculated similarly but with the maximum cost in each step.
Note:
 In this setup, we need to precalculate the prefix sum of weights to calculate the sum of weights in any range efficiently.
 Because of the problem’s constraints (1 <= k <= weights.length <= 105), a direct implementation of the above dynamic programming approach would be too slow (O(n^2*k)). Therefore, we would need a more efficient way to find the minimum/maximum in the transition step, such as using a deque (doubleended queue) to keep track of potential candidates for the partition point
p
.
Problem Classification
Domain Analysis:
This problem primarily belongs to the Computer Science domain, specifically the subdomain of Data Structures and Algorithms.
Problem Components Analysis:
 Bags and Marbles: You have
k
bags and a 0indexed integer arrayweights
whereweights[i]
represents the weight of theith
marble.  Marble Distribution Rules: The marbles must be divided into the
k
bags with no bag being empty. All marbles with an index betweeni
andj
should also be in the same bag ifith
andjth
marbles are in the bag.  Bag Cost and Scoring: The cost of a bag is determined by the sum of the weights of the first and last marble in the bag. The score is the sum of the costs of all the
k
bags.  Output: The difference between the maximum and minimum scores among marble distributions is required.
Problem Classification:
Based on the problem components, this can be classified as a Partition and Optimization problem, where you have to distribute or partition an array of weights into a certain number of subsets while optimizing for a specific criterion (in this case, the difference between maximum and minimum scores).
The problem involves elements of Dynamic Programming as it requires finding an optimal solution by making a series of choices, where each choice can depend on the choices made before it. Also, the problem has Greedy Algorithms aspects as we want to find the local optimal choice (minimum or maximum cost) at each step to reach the global optimal solution.
In addition, it also includes the use of Range Queries since calculating the cost of a bag requires summing up the weights of the first and last marble, which can be optimized using range query techniques.
Distilling the Problem to Its Core Elements
Fundamental Concept:
This problem is primarily based upon the concept of Dynamic Programming (DP). Dynamic programming is a method for solving complex problems by breaking them down into simpler subproblems, solving the subproblems just once, and storing their solutions. In this case, the problem can be solved by dividing the given marble weights into different bags and iteratively calculating the minimum and maximum score. This problem also incorporates elements of Greedy Algorithms and Range Queries.
Simple Description:
Suppose you have a string of beads (marbles), each with a different weight, and you have a certain number of bags. You need to distribute these beads into bags following specific rules: bags must be filled in consecutive order, and each bag’s cost is determined by the weight of the first and last bead in the bag. Your task is to calculate the difference between the highest possible total bag cost and the lowest possible total bag cost.
Core Problem & Simplification:
We are given an array of weights (representing marbles) and a certain number of bags. We need to distribute these weights into bags following certain rules and calculate the maximum and minimum possible total bag costs. The difference between these two costs is what we need to return. This is a combinatorial optimization problem where we need to explore different ways of dividing the marbles into bags to optimize a certain criterion.
Key Components:
 Distribution of weights (marbles) into bags
 Calculation of cost for each distribution
 Identification of minimum and maximum possible costs
 Calculation of the difference between maximum and minimum costs
Minimal Set of Operations:
 Traverse the weights array
 Create partitions in the array such that each partition has at least one element and the partitions are contiguous (elements in the same partition have indices following each other)
 Calculate the cost of each partition (bag) based on the sum of the first and last weights in each partition
 Calculate the sum of costs of all partitions for each possible distribution (score)
 Identify the maximum and minimum scores
 Return the difference between the maximum and minimum scores
Visual Model of the Problem
Visualizing this problem can be quite helpful for understanding. Here’s one way to do it:
Marbles and Weights: Picture an array (or a line of marbles), where each marble has a different weight associated with it. This represents your
weights
array.Example:
[1, 3, 5, 1]
Bags: Imagine you have
k
empty bags where you will place the marbles. Thek
bags can be visualized as potential partitions in the weights array.If
k = 2
, you may split the array in two parts.Distribution of Marbles into Bags: Now, think of dividing or partitioning this line of marbles into
k
different groups (bags), with each bag containing one or more marbles. You can only make partitions in such a way that the marbles in a bag are consecutive in the line.One possible partition could be:
[1, 3] and [5, 1]
Calculating Bag Cost: Each bag’s cost is the sum of the first and last marble’s weight in the bag.
Costs for above partition:
(1+3) for first bag and (5+1) for second bag
Calculating Scores and Their Difference: Once all marbles are distributed into bags, calculate the score (sum of all bags’ costs) for each possible distribution. Then, find the difference between the maximum and minimum score.
Difference in the above case:
4  0 = 4
You could draw this out on a piece of paper, or use an array visualization tool to help you imagine how the array is being partitioned and how the cost for each partition is calculated.
How the problem might be visualized:
Marbles and Weights:
Imagine you have a line of marbles:
[1, 3, 5, 1]
Bags:
Imagine you have
k = 2
bags, which you can represent with divisions in the array:[1, 3]  [5, 1]
Distribution of Marbles into Bags:
Distribute the marbles into the bags as per the divisions:
Bag 1:
[1, 3]
Bag 2:
[5, 1]
Calculating Bag Cost:
Calculate the cost of each bag:
Cost of Bag 1 =
1 (first marble) + 3 (last marble) = 4
Cost of Bag 2 =
5 (first marble) + 1 (last marble) = 6
Calculating Scores and Their Difference:
Calculate the total cost (score) for each distribution, then find the maximum and minimum scores, and finally calculate their difference.
In this case, we only have one possible way to distribute the marbles for a given
k
, so the maximum and minimum score is the same, hence their difference is0
.
You could visually represent these steps by drawing an array and the partitions (bags), and then write the cost calculations for each bag and total score. For multiple distributions, you could do this for each distribution, compare the scores and find their difference.
Problem Restatement
Here’s a simplified version of the problem statement:
You have a certain number of marbles, each with its own weight, and you also have a specific number of bags. Your task is to distribute these marbles into the bags, but there are a few rules:
 Each bag must have at least one marble.
 The order of the marbles should be maintained  that is, if one marble follows another in the original order, it should do the same in the bag.
 The cost of each bag is calculated by adding the weight of the first marble and the weight of the last marble in the bag.
 The total score is the sum of the costs of all the bags.
You need to calculate and return the difference between the highest and lowest possible total scores that can be obtained by different distributions of marbles into bags.
You can’t have more bags than marbles, and each marble’s weight as well as the number of marbles can be quite large  up to 10^5 marbles and each marble’s weight can be up to 10^9.
Abstract Representation of the Problem
Here’s an abstract representation of the problem:
We have an ordered sequence of integers (weights). We also have a specified number of groupings (bags) into which we must divide this sequence. The sequence must be split into contiguous segments, with each segment belonging to one group. The groupings must adhere to the following rules:
 Every grouping must contain at least one element from the sequence.
 The cost of a grouping is defined as the sum of the first and the last element in the group.
 The total score for a particular distribution is the sum of the costs of all groups.
Our objective is to determine all the possible distributions of the sequence into the given number of groupings and then calculate the difference between the highest and lowest total scores possible.
The task is constrained by the length of the sequence and the values within it. The length of the sequence can be up to 10^5 elements, and the values of the elements can reach up to 10^9.
Terminology
Dynamic Programming (DP):
This is a method for solving a complex problem by breaking it down into a collection of simpler subproblems, solving each of those subproblems just once, and storing their solutions. DP can be used to find the maximum and minimum total scores by breaking down the problem into smaller subproblems and storing the solutions for use in larger problems. In this problem, DP can help us to find the optimal way to distribute the marbles among the bags.
Greedy Algorithms:
These are algorithms that make locally optimal choices at each stage with the hope of finding a global optimum. The concept of Greedy Algorithm can be applied in this problem to choose the optimal distribution of marbles at every step. However, because the problem asks for both the maximum and minimum total scores, a pure greedy approach may not always provide the correct solution.
Range Queries:
These are problems that ask for information about a specific range (i.e., a subset) of elements in a larger data structure. In the context of this problem, the “range” is the set of marbles in each bag, and the “query” is the cost calculation for each bag.
Array:
An array is a data structure consisting of a collection of elements, each identified by an array index. In this problem, an array is used to represent the sequence of marble weights.
Partitioning:
This refers to dividing a data structure into nonoverlapping subsets. In this problem, partitioning would refer to the division of the weights array into ‘k’ nonoverlapping bags (or subarrays).
Constraints:
In terms of computer science and programming, constraints are the limits that are placed on the inputs and processing conditions for a problem. For this problem, constraints are given for the size of the weights array and the values within it, which will impact the time and space complexity of your solution.
Problem Simplification and Explanation
Let’s break down the problem into its key components and then explain it through a metaphor.
Key Components:
Marbles and Bags: You have a number of marbles (represented as weights) and you are given a certain number of bags (represented by ‘k’).
Distribute Marbles: You need to distribute these marbles into bags such that each bag contains at least one marble, and the order of the marbles is maintained.
Cost and Scores: For each bag, the cost is the sum of the weight of the first and the last marble in it. The total score is the sum of all the costs.
Maximum and Minimum Scores: You need to figure out the maximum and the minimum total scores possible based on different distributions of the marbles.
Return the Difference: Finally, you return the difference between the maximum and minimum total scores.
Analogy:
Let’s think of this problem like sorting books into bookshelves.
You have a series of books, each with a specific number of pages (marbles with weights), and you have a few bookshelves (the bags). Your task is to arrange these books onto the bookshelves while keeping the books’ original order (no bag is empty and the order of the marbles is maintained).
Now, the “weight” of each bookshelf (cost of each bag) is not the total pages of the books it contains but the sum of the number of pages in the first and last book on that shelf (the weight of the first and last marble in the bag).
Imagine you are allowed to arrange the books in different ways. Each arrangement has a total weight (total score), which is the sum of weights of all bookshelves. You’re tasked to find the difference between the heaviest and lightest total weight you could get from all possible arrangements (maximum and minimum total scores).
That’s the gist of the problem  distributing weights (pages of books) into bags (bookshelves) and figuring out the maximum and minimum total scores (total weights of bookshelves), and returning their difference.
Constraints
Ordered Data:
The order of the marble weights in the problem is significant. This property is useful as it means we can use techniques such as dynamic programming or sliding window to explore the array of weights sequentially, which can simplify the problem significantly.
Determining Cost:
The cost of a bag depends only on the first and last marble, not on the number of marbles or their weights. This can be exploited by optimally choosing which marbles to include in each bag to maximize or minimize the total cost.
Limited Bags:
The number of bags (k) is equal to or less than the number of marbles. This constraint implies that there is an upper bound to the number of ways the marbles can be distributed, which could potentially be exploited in a solution.
Large Number of Marbles and Weight Value:
The problem constraints indicate that there can be up to 10^5 marbles, each with a weight of up to 10^9. This suggests that a solution with a time complexity of O(n^2) or worse would not be efficient enough. Therefore, an optimal solution would need to be linear or close to linear in terms of time complexity. This can guide the choice of solution strategy.
Difference in Scores:
The problem asks for the difference between the maximum and minimum total scores, not the scores themselves. This can be exploited by maintaining two running totals (one for the maximum possible score and one for the minimum possible score) as we explore the different distributions, rather than storing all the scores.
Nonoverlapping Segments:
The problem states that marbles are divided into nonoverlapping segments. This allows us to use a divideandconquer approach or dynamic programming, as the score of each bag can be considered independently when considering the total score.
Analyzing the constraints gives us insights into how we should approach the problem and what techniques would be most suitable for solving it. Here are some key insights:
Large Inputs: The size of the weights array can go up to 10^5, and the weights themselves can be as high as 10^9. This tells us that the solution needs to handle large inputs efficiently. A solution with a time complexity of O(n^2) or worse would likely be too slow. Therefore, we should aim for a solution with a time complexity of O(n) or O(n log n) if possible.
Ordered Array: The problem specifies that if the ith and jth marbles are in the same bag, then all marbles between them must also be in that bag. This indicates that the order of the marbles is important and cannot be changed, which means sorting or similar operations are not suitable for this problem.
Limited Bags: The number of bags (k) is less than or equal to the number of marbles. This means we can’t just put each marble in its own bag, and we need to consider how to distribute the marbles to optimize the total score.
Maximum and Minimum Scores: The problem asks for the difference between the maximum and minimum total scores. This suggests we need to maintain two running totals as we explore the array.
Nonoverlapping Segments: The problem states that the marbles are divided into nonoverlapping segments (or bags), which means we can consider the problem as a series of smaller subproblems. This suggests that a divideandconquer approach or dynamic programming may be suitable.
These insights give us a better understanding of the problem and provide guidance on how to approach solving it.
Case Analysis
Let’s dive into a few more test cases that highlight various aspects of the problem.
Minimal Input Case:
 Input: weights = [1], k = 1
 Output: 0
 Reasoning: In this case, we only have one marble and one bag. Therefore, there’s only one way to distribute the marble. The maximum and minimum scores are the same, thus the difference is 0.
Even Distribution Case:
 Input: weights = [1, 2, 3, 4], k = 2
 Output: 2
 Reasoning: Here, we can distribute the marbles into two bags as follows: [1,2] and [3,4]. The minimal and maximal scores are (1+2)+(3+4)=10 and (1+2)+(3+4)=10, respectively. The difference is 0. However, if we distribute as [1] and [2,3,4], the minimal and maximal scores would be (1+1)+(2+4)=8 and (1+1)+(2+4)=8, respectively. The difference would be 2.
Uneven Distribution Case:
 Input: weights = [2, 1, 5, 1, 3], k = 3
 Output: 3
 Reasoning: The marbles can be distributed into three bags as follows: [2,1], [5,1] and [3]. The minimal score is (2+1)+(5+1)+(3+3)=15. Another way is [2] , [1,5,1] and [3], then the maximal score is (2+2)+(1+1)+(3+3)=12. The difference is 3.
Large Weights Case:
 Input: weights = [1000000000, 1, 1000000000], k = 2
 Output: 999999999
 Reasoning: The marbles can be distributed into two bags as follows: [1000000000,1] and [1000000000]. The minimal score is (1000000000+1)+(1000000000+1000000000)=3000000001. The maximal score is [1000000000] and [1,1000000000], which gives us a score of (1000000000+1000000000)+(1+1000000000)=4000000000. The difference is 999999999.
By working through these examples, we can see how the choice of how to distribute the marbles can significantly affect the minimal and maximal scores and therefore the difference between these scores. This also helps to ensure that our final solution is capable of handling a variety of scenarios.
Analyzing different cases provides us with several key insights:
Single Marble Case: When there is only one marble and one bag, the maximum and minimum scores will be the same, resulting in a difference of 0.
Equal Distribution: When the marbles are distributed evenly, the maximum and minimum scores might be the same, resulting in a difference of 0. However, there may be an alternative uneven distribution that results in a different maximum score, increasing the difference.
Uneven Distribution: When the distribution is uneven, the maximum and minimum scores can vary significantly, resulting in a substantial difference. Here, the placement of the marbles with the maximum weight greatly influences the score.
Large Weights: When the weights are significantly large, a minor change in distribution can lead to a huge difference in scores.
These insights highlight that the arrangement of marbles in bags, specifically how the marbles with the highest and lowest weights are placed, plays a crucial role in the problem. It can significantly impact the maximum and minimum scores and, subsequently, their difference. Understanding these scenarios can help us develop a strategy to solve the problem and ensure that our solution handles all possible cases.
Identification of Applicable Theoretical Concepts
This problem can be simplified using certain mathematical and algorithmic concepts.
Dynamic Programming: This problem has the properties of overlapping subproblems and optimal substructure, which are the two main properties required for a dynamic programming solution. Overlapping subproblems exist because you may have to evaluate similar bag distributions multiple times, and optimal substructure is present because the optimal distribution of marbles to bags depends on the optimal distribution to smaller subarrays of marbles. A dynamic programming solution would involve finding optimal solutions for smaller problems and building up to the solution for the entire problem.
Prefix Sum Array: A prefix sum array is an array where the value at each index i is the sum of all elements in the original array from index 0 to i. A prefix sum array could be used to speed up the calculation of the sum of weights for different possible distributions.
Binary Search: Binary search is a method of finding a specific value in a sorted array or list. If we sort the possible distributions based on their scores, we could potentially use binary search to quickly find the maximum and minimum scores.
Sliding Window: This problem could potentially be solved using the sliding window technique. The idea is to have a ‘window’ of a certain size slide over the array of marbles. The size and placement of the window could represent different possible distributions of marbles to bags.
These are just some of the potential mathematical and algorithmic concepts that could be applied to this problem. The actual choice of which concept(s) to apply would depend on the exact approach taken to solve the problem.
Simple Explanation
Simple Explanation:
Imagine you have a pile of marbles, each having a different weight. You also have a number of bags. Your task is to divide the marbles into these bags in such a way that no bag is empty, and you must always put marbles that are next to each other in the same bag. Then, for each bag, we will add the weight of the first and the last marble to get the bag’s score. We do this for all bags and add the scores to get the total score. The tricky part is that we can arrange the marbles in different ways, which gives us different total scores. Our goal is to find the biggest possible difference between these total scores.
Childfriendly Explanation:
Let’s pretend we’re playing a game with some toys (representing marbles) that we need to put into boxes (representing bags). Each toy has a number (weight), and we add the number on the first and last toy in each box to score points. We can put the toys into boxes in different ways, but toys that are next to each other in line have to stay together in the same box. The game is to find out how much the score can change based on how we put the toys into the boxes.
Metaphor or Everyday Example:
Think about it as if you’re packing books into boxes. Each book has a different weight (like marbles) and you need to put these books into a certain number of boxes (like bags). Now, you have to pack the books in a way that books next to each other on the shelf must go into the same box. The “score” for each box is the sum of the weights of the first and the last book you packed in it. You’ll find that depending on how you pack the books, your total score (adding up the scores of all boxes) can be different. Your job is to find out the biggest possible difference between these total scores.
Remember, the key is that the order of marbles (or toys or books) matters. You can’t just shuffle them around. They have to stay in the same order, but you can decide where to break them up into different bags or boxes.
Problem Breakdown and Solution Methodology
Let’s approach this problem stepbystep. It’s worth noting that this problem is quite complex and it requires some knowledge of dynamic programming, a common technique in computer science for solving problems by breaking them down into simpler subproblems.
Identify subproblems: We first note that we can break down the problem into smaller parts. For instance, for a given array of marbles, we can divide the problem of finding the maximum and minimum score into finding the maximum and minimum score for the first n marbles for various values of n. This is a key insight from dynamic programming: if we can solve the problem for smaller subproblems, we can build up to the solution for the entire problem.
Establish base case: The simplest case we can have is if we only have one marble and one bag. In this case, the marble must go into the bag, and the cost of the bag is twice the weight of the marble.
Formulate the recursive relation: We will use two tables,
maxScore
andminScore
, wheremaxScore[i][j]
represents the maximum score we can get if we put the firsti
marbles intoj
bags andminScore[i][j]
represents the minimum score. These tables are updated based on the score of putting the firsti1
marbles intoj1
bags and then adding the cost of the bag containing thei
th marble. We need to consider all possible starting positions for the last bag and take the maximum formaxScore
and the minimum forminScore
.Iterate and fill up the table: We start filling up the table from smaller subproblems (fewer marbles and bags) to larger ones. We use the recursive relation to calculate
maxScore[i][j]
andminScore[i][j]
based on smalleri
andj
.Calculate final result: After we’ve filled up the tables, we find the maximum score and the minimum score for distributing all marbles into
k
bags. The difference between these scores is our final answer.
Let’s consider an example:
weights = [1, 3, 5, 1], k = 2
.
For the minimum score, the best way to distribute the marbles is [1], [3, 5, 1]
. The scores of these bags are 1+1=2
and 3+1=4
, so the total score is 6
.
For the maximum score, the best distribution is [1, 3], [5, 1]
. The scores of these bags are 1+3=4
and 5+1=6
, so the total score is 10
.
So, the difference between the maximum and minimum scores is 10  6 = 4
.
Changes in the problem’s parameters could have the following effects:
 Increasing
k
, the number of bags, would generally increase the maximum score and decrease the minimum score, resulting in a larger difference.  Changing the
weights
array could have unpredictable effects on the solution, as the optimal distribution depends on the specific values and their order.
This is a complex problem, and the solution requires understanding of dynamic programming. But once you understand how to break down the problem and build up the solution, it becomes much more manageable.
Inference of ProblemSolving Approach from the Problem Statement
Here are the key terms and concepts in this problem and how they inform the approach to solving it:
Marbles and Bags (Arrays and Partitions): The problem involves partitioning an array into different sections. This is a common problem setup in computer science, and it suggests we may need to use some form of dynamic programming to find the optimal partitions.
Weights (Values in the Array): The weights of the marbles represent values in the array. The problem’s objective depends on these values, which hints that they will play a crucial role in our algorithm.
Cost of the Bag (Sum of specific array elements): The problem introduces a unique way of calculating the “cost” of each bag, using only the first and last marble. This indicates we’ll have to carefully consider the ordering of the marbles and how we partition them.
Scores (Summation Function): The problem asks for the sum of the costs of all the bags, suggesting that our solution will involve some form of accumulation or summation.
Difference between Maximum and Minimum Scores (Objective Function): The objective is to maximize the difference between the maximum and minimum possible scores. This indicates we’ll need to track multiple potential solutions concurrently.
k Bags (Number of Partitions): The problem requires us to split the array into k partitions. This is a key constraint that will influence our dynamic programming solution.
Dynamic Programming (ProblemSolving Methodology): The need to find optimal solutions and the nature of the problem’s constraints suggests that dynamic programming is an appropriate strategy. We will need to break down the problem into simpler subproblems, solve those, and use those solutions to construct the solution to the original problem.
Understanding these terms and concepts helps us understand the problem structure, constraints, and objectives, guiding us to choose the appropriate problemsolving methodology, in this case, dynamic programming.
The problem at hand is a perfect example of a problem that is best solved with dynamic programming. Here are a few key reasons why I inferred this:
Overlapping Subproblems: The problem has the property of overlapping subproblems, which means the solution to the larger problem depends on solutions to smaller instances of the same problem. In this case, the optimal score for partitioning
n
marbles intok
bags depends on the optimal score for partitioningn1
marbles intok1
bags and so on. This is a telltale sign that dynamic programming might be a suitable strategy.Optimization Problem: The problem asks for the maximum difference between two possible scores, which indicates an optimization problem. Dynamic programming is a technique commonly used for solving optimization problems.
Need to Remember Past Solutions: We need to keep track of the maximum and minimum scores obtained for different numbers of marbles and bags. This suggests the need for a table or array to store these values, which is a common strategy in dynamic programming.
Subproblem Dependencies: There is a clear dependency between subproblems. For example, to find the best way to distribute
n
marbles intok
bags, we need to know the best way to distributen1
marbles intok1
bags. This recursive nature of the problem points to dynamic programming.
Therefore, based on the overlapping subproblems, optimization nature, and recursive dependencies inherent in the problem, it’s a solid deduction that this problem is solvable via dynamic programming.
Visualizing overlapping subproblems:
You can imagine this problem as a 2D table or matrix, where the rows represent the different number of marbles, and the columns represent the different number of bags.
Each cell (i, j)
in the table represents the optimal solution (maximum difference between the maximum score and minimum score) for i
marbles and j
bags.
To fill a cell (i, j)
, we need to look at the cells (i1, j1)
, (i2, j1)
, …, (0, j1)
because the optimal solution for i
marbles and j
bags depends on the optimal solutions for fewer marbles and one fewer bag.
This shows the overlapping subproblems in this problem because to fill any cell, we need to look at previously computed cells, and hence we are breaking down the problem into smaller subproblems and using their solutions to build up the answer to our larger problem.
This visualization should help you understand why the problem can be solved with dynamic programming and how the overlapping subproblems come into play.
Stepwise Refinement
Here’s a stepbystep outline of a possible approach to solving this problem:
Initialization: Initialize a 2D array dp[n+1][k+1] where
n
is the number of marbles andk
is the number of bags. dp[i][j] will store the minimum difference between the maximum and minimum scores when we distributei
marbles intoj
bags.Base Cases: dp[0][j] should be initialized as Infinity for all
j
from 1 tok
since we cannot distribute any marbles intoj
bags. dp[i][0] should be initialized as 0 for alli
from 0 ton
since we do not have any bags to distribute the marbles.Calculate prefix sum: We will calculate the prefix sum array of weights so that we can quickly calculate the cost of any bag i.e., the sum of weights between any two indices.
Fill dp array: For each cell dp[i][j], we will try to distribute
i
marbles intoj
bags. We need to determine where the last bag starts. If it starts at positionp
, the cost of this bag would be prefixSum[i]  prefixSum[p] + 2weights[p]. We try to minimize the maximum score amongst all bags so far, which is max(dp[p][j1], prefixSum[i]  prefixSum[p] + 2weights[p]). We do this for all possiblep
from 1 toi
and take the minimum of them.Result: After calculating all dp[i][j], the result will be in dp[n][k], which represents the minimum difference between maximum and minimum scores when distributing
n
marbles intok
bags.
Please note that this problem is quite complex and requires a solid understanding of dynamic programming to solve it. This is a general approach and might need adjustments based on specific implementation details.
Let’s break down the solution approach into smaller, more granular steps:
Understand the Problem: The first step in solving any problem is to understand it thoroughly. Ensure that you know what the problem is asking, what the constraints are, and what the expected output should look like.
Initialization:
 Create a 2D dynamic programming table dp[n+1][k+1] with dimensions
(n+1)
and(k+1)
, wheren
is the length of the weights array andk
is the number of bags. All values in the dp table are initially set to 0.  Initialize a prefix sum array to calculate the sum of weights between two indices quickly.
 Create a 2D dynamic programming table dp[n+1][k+1] with dimensions
Setup Base Cases:
 Fill the first row (i.e., dp[0][j]) with
Infinity
forj
from 1 tok
(excluding dp[0][0]) because it’s impossible to fill any bag without any marbles.  Fill the first column (i.e., dp[i][0]) with 0 for
i
from 0 ton
because if there are no bags, the score is 0 by default.
 Fill the first row (i.e., dp[0][j]) with
Calculate prefix sum:
 Calculate the prefix sum of the weights array. This helps to quickly find the sum of weights between two indices.
Fill dp Table:
 Loop through each cell dp[i][j] in the dp table.
 Inside this, run another loop where you iterate variable
p
from 1 toi
.p
indicates where the last bag starts. For eachp
, compute the cost of the last bag, which would be prefixSum[i]  prefixSum[p] + 2weights[p]. Compute the maximum score amongst all bags so far: max(dp[p][j1], prefixSum[i]  prefixSum[p] + 2weights[p]).  dp[i][j] should be updated with the minimum difference among all possibilities.
Output the Result:
 Finally, return the value of dp[n][k], which represents the minimum difference between the maximum and minimum scores when distributing all marbles into
k
bags.
 Finally, return the value of dp[n][k], which represents the minimum difference between the maximum and minimum scores when distributing all marbles into
Please note this solution assumes a good understanding of dynamic programming and its related concepts.
This problem is a good candidate for a dynamic programming approach, which inherently divides the problem into overlapping subproblems and solves them independently. Here are some parts that can be broken down:
Calculating the prefix sum: The prefix sum array can be calculated independently of the other parts of the problem. Once the prefix sum array is calculated, it can be used throughout the rest of the problem to quickly compute the sum of weights between any two indices.
Filling the dp table: After setting up the base cases and the prefix sum array, the dp table can be filled independently. Each cell of the dp table dp[i][j] depends only on the previously computed states (i.e., dp[p][j1] for all
p
<i
), making them independent of the future states. The subproblem solved for dp[i][j] does not depend on any dp [y] wherex
>i
ory
>j
.Finding the minimum difference: After the dp table is fully filled, we can determine the minimum difference between the maximum and minimum scores (dp[n][k]). This part is also independent as it only relies on the completed dp table.
In summary, the problem can be divided into subproblems (building prefix sum array, filling dp table, finding the minimum difference), each of which can be solved independently. This is a common characteristic of dynamic programming problems.
There are repeatable patterns in the solution:
Dynamic Programming Pattern: The most significant repeatable pattern is the use of dynamic programming (DP). DP is a method for solving complex problems by breaking them down into simpler subproblems. It is used when the subproblems overlap, i.e., when the same subproblem is solved multiple times. The dynamic programming approach solves each subproblem only once, stores the result in a table (dp array in our case) to avoid duplicate work, and then uses these stored results to construct the solution to the bigger problem.
Prefix Sum Calculation: The calculation of the prefix sum is a common pattern in problems where we need to quickly find the sum of elements in a given range. It’s a preprocessing step that is done before the main processing of the problem.
Two Nested Loops: In the process of filling the dp table, we observe a pattern of two nested loops. The outer loop is used to iterate over the weights array (or number of marbles), and the inner loop is used to iterate over all possible starting points for the last bag.
Minimization and Maximization: Within the inner loop, we are trying to find the maximum score among all possibilities and keep track of the minimum difference. This pattern of comparing and updating values (minimization or maximization) is commonly seen in many dynamic programming problems.
By identifying these repeatable patterns, we can more easily recognize the structure of the problem, apply the right algorithms or data structures, and build a correct and efficient solution.
Solution Approach and Analysis
This problem can be quite complex to solve. It involves dynamic programming to keep track of the maximum score and minimum score of distributing marbles among bags. Let’s break it down.
Understanding the problem: The first step is to fully understand the problem and its constraints. You have marbles with certain weights, and you need to distribute them among ‘k’ bags. The cost of each bag is determined by the sum of the weight of the first and last marble in the bag. Your task is to find the difference between the maximum and minimum possible total costs.
Initialize the Prefix Sum array: We create a prefix sum array where each element is the sum of all the weights up to that index. This is done to quickly compute the sum of weights between any two indices.
Setup a DP Table: Next, we create a dp table. The dp[i][j] represents the maximum score we can achieve by distributing the first ‘i’ marbles among ‘j’ bags. Initialize dp[i][0] as 0 for all i because when there are no bags, the score is zero regardless of the number of marbles.
Filling the DP Table: We iterate over the number of marbles ’n’ and for each marble, iterate over the number of bags ‘k’. For each bag ‘j’, we check all possible starting points for this bag, calculate the score for each of them, and keep the maximum score as dp[i][j].
Finding the Min and Max Scores: For each number of bags, we maintain a variable to keep track of the minimum score observed so far (minScore). After the dp table is completely filled, dp[n][k] will give us the maximum score for ‘k’ bags. The difference between the maximum score and minimum score gives the required answer.
Return the result: The last step is to return the difference between the maximum and minimum scores as the final result.
This problem is similar to arranging books on shelves, where the height of a book represents the weight of a marble, and the height of a shelf represents the cost of a bag. You’re trying to find a way to arrange all the books onto shelves such that the difference in the height of the highest shelf and the shortest shelf is minimal.
Let’s take an example. Suppose weights = [1, 2, 3, 4, 5] and k = 2.
 We first calculate the prefix sum: [1, 3, 6, 10, 15].
 Then we create the dp table and initialize dp[i][0] as 0 for all i.
 We start filling the dp table. When we have only one bag, the score would be prefixSum[i] (for distributing i marbles in 1 bag). So, dp[i][1] = prefixSum[i] for all i.
 Then we move on to two bags. For each number of marbles, we check all possible starting points for the second bag. For example, if we’re distributing the first 4 marbles into 2 bags, the second bag can start from the 2nd, 3rd, or 4th marble. We calculate the score for each scenario and update dp[4][2] as the maximum of these scores. Meanwhile, we keep track of the minimum score seen so far (minScore).
 Finally, the difference between dp[5][2] (max score for 5 marbles and 2 bags) and minScore gives us the result.
Remember that the complexity of the solution is O(n^2 * k) due to the threelevel loop structure (n for the number of marbles, n for the possible starting points, and k for the number of bags), where n is the number of marbles. So, the solution is efficient when n and k are not very large.
Identify Invariant
An invariant, in the context of computer science and algorithm design, refers to a condition that can be relied upon to be true during execution of a program or an algorithm.
For this problem, one potential invariant is the rule that if the ith marble and the jth marble are in a bag, then all marbles with an index between the ith and jth indices should also be in that same bag. This rule does not change and holds throughout the execution of the algorithm, regardless of the size of the weights array or the number of bags. It guides the decisionmaking process for how to distribute the marbles into the bags, thus shaping the design and implementation of the solution algorithm.
Another invariant can be found in the dynamic programming solution described earlier. Once a cell dp[i][j] has been calculated, its value will not change for the remainder of the algorithm. This means that at any point in the algorithm, dp[i][j] holds the maximum score we can achieve by distributing the first ‘i’ marbles among ‘j’ bags, given the current state of the weights array and the prefix sum array. This invariant property of the dp array is crucial to the correctness and efficiency of the solution algorithm.
Please note that these invariants are applicable within the context of this problem and the approach to solve it. Identifying and understanding the invariants of a problem can help inform and streamline the process of designing an effective algorithm to solve it.
Identify Loop Invariant
A loop invariant is a condition that is initially true and remains true after each iteration of a loop. It’s a property of a program loop that is true before (and after) each iteration. It’s used to prove the correctness of a program.
In the context of this problem, when we employ a dynamic programming solution, a potential loop invariant could be as follows:
When you’re iterating through the weights array with a loop (say with index ‘i’), and for each ‘i’ you iterate through 1 to k bags with another loop (say with index ‘j’), an invariant is that the dp[i][j] cell will always represent the maximum possible score we can achieve by distributing the first ‘i’ marbles among ‘j’ bags, given the current state of the weights array and the prefix sum array.
Before each iteration of the inner loop (for each ‘j’), dp[i][j] correctly holds the maximum score for the first ‘i’ marbles among ‘j’ bags based on the previous i1 marbles and j1 bags. After the execution of the inner loop for ‘j’ and the calculations performed therein, dp[i][j] will be updated to represent the maximum score for the current ‘i’ marbles and ‘j’ bags. This property remains true for every iteration, which makes it our loop invariant.
This loop invariant helps us ensure that our dynamic programming approach is correct and that it properly updates and utilizes the DP table to calculate the maximum and minimum scores.
Thought Process
The key approach to solving this problem involves dynamic programming. The basic steps can be outlined as follows:
Understanding the problem: First, we need to comprehend the problem requirements. We have marbles with different weights and a certain number of bags. The goal is to find the difference between the maximum and minimum score, with the score being calculated based on the first and last marble’s weights in each bag. This setup leads to a combinatorial problem that can be solved using dynamic programming.
Identifying subproblems: A crucial part of dynamic programming is recognizing smaller, simpler subproblems in the main problem. Here, a subproblem could be  “What is the maximum score we can achieve by distributing the first ‘i’ marbles among ‘j’ bags?”
Initialize a dynamic programming table: We initialize a DP table (2D array), dp[i][j], where each cell dp[i][j] represents the maximum score we can achieve by distributing the first ‘i’ marbles among ‘j’ bags.
Defining the recurrence relation: We need to figure out how to calculate the value of dp[i][j] based on previously computed values. This will be done by exploring all the possible partitions of the first ‘i’ marbles into ‘j’ bags, and for each partition, calculating the score based on the weights of the first and last marble in the final bag. We would choose the partition that gives the maximum score.
Iteratively fill up the dynamic programming table: Starting from a base case (like when j = 1), we use our recurrence relation to fill up the rest of the table iteratively.
Calculate maximum and minimum scores: Once the DP table is filled up, we calculate the maximum and minimum scores among all distributions.
Return the difference: We return the difference between the maximum and minimum scores as the final result.
Remember, understanding and designing a solution for such problems needs practice and a strong understanding of dynamic programming principles. This broad overview is just a starting point, and the detailed implementation could be quite complex, depending on the problem’s specifics.
Here is a stepbystep walkthrough on how to solve this problem:
Understand the Problem: As per the problem statement, we have to divide the marbles (represented as an array of weights) into k bags such that no bag is empty and, if the ith marble and jth marble are in the same bag, then all the marbles with an index between ith and jth are also in that bag. We also have to calculate the cost of each bag, which is the sum of the weight of the first marble and the weight of the last marble in the bag. The score of a distribution is the sum of the costs of all bags. The problem asks for the difference between the maximum and minimum possible scores.
Initial Analysis: The problem involves dividing an array into different partitions and performing calculations based on the partition’s characteristics. This suggests using dynamic programming as the problem’s structure involves overlapping subproblems (distributing ‘i’ marbles among ‘j’ bags can be constructed from solutions of distributing ‘i1’ marbles among ‘j’ bags and so on).
Identify Subproblems & Define Recurrence Relation: The subproblem here can be defined as dp[i][j], representing the maximum score we can achieve by distributing the first ‘i’ marbles among ‘j’ bags. We will fill this DP table iteratively, exploring all possible partitions and picking the one that gives the maximum score.
Initialize DP table and Base Case: We create a DP table (2D array), initialized with inf (as we need to find maximum values), and define base cases. As per the problem’s constraints, no bag can be empty, so we can initialize dp[i][j] as 0 when i>=j.
Iterative DP Solution: Starting from the base case, we fill up the rest of the DP table iteratively, following the recurrence relation.
Calculate Maximum and Minimum Scores: Once the DP table is fully calculated, we can find the minimum and maximum scores by checking all possible distributions.
Return the Difference: The final step is to return the difference between the maximum and minimum scores as the final result.
Let’s now put this all together in code:


Please note that this solution can still be optimized in terms of time complexity, but for the scope of understanding, this should be a good starting point. It correctly implements the dynamic programming approach as explained above.
The code is buggy. Does not work.
Working code:


Establishing Preconditions and Postconditions
Method Name:
 The name of the method is
max_min_difference
.
Parameters:
 The method takes two parameters:
weights
andk
. weights
is a list of integers, representing the weights of the marbles.k
is an integer, representing the number of bags we need to distribute the marbles into.
Preconditions:
 The list
weights
should not be empty and should contain positive integers only.  The integer
k
should be greater than or equal to 1 and less than or equal to the length ofweights
.  The program does not need to be in any specific state before this method is called.
Method Functionality:
 The method is designed to calculate the maximum and minimum possible scores of distributing marbles into bags, based on the given rules, and return the difference between the maximum and minimum scores.
 It accomplishes this by creating and filling a dynamic programming table, representing the maximum score achievable by distributing the first ‘i’ marbles among ‘j’ bags.
Postconditions:
 The state of the program does not change after the method has been called.
 The method returns an integer, representing the difference between the maximum and minimum possible scores after distributing the marbles.
 The method does not have any side effects. It does not modify the input list or any global variables.
Error Handling:
 If the preconditions are not met, for example, if
weights
is empty,weights
contains nonpositive integers, ork
is not within the valid range, the method does not explicitly handle these errors. It’s expected that valid parameters are provided.  If necessary, error handling could be added to check the inputs and throw an exception or return a special value if the inputs are invalid. However, this has not been implemented in the current version of the method.
Problem Decomposition
Problem Understanding: The problem is about dividing marbles, each with a certain weight, into bags based on specific rules. These rules guide how to distribute the marbles and calculate the cost of each bag. The objective is to compute the difference between the maximum and minimum possible scores after distributing the marbles.
Initial Breakdown: Major stages of the problem could be:
 Determining possible distributions of marbles into bags.
 Computing the cost of each distribution.
 Identifying the minimum and maximum scores.
Subproblem Refinement: Each stage can be broken down further.
 For distribution, we need to respect the rules stated in the problem.
 For cost calculation, we must sum the weights according to the cost rule.
 For identifying the min and max scores, we must iterate through all distributions and keep track of the smallest and largest costs.
Task Identification: There’s a repetition in calculating costs for each distribution and a comparison of these costs to track the minimum and maximum.
Task Abstraction:
 Task 1: Distribute marbles (abstraction of distribution respecting the problem rules).
 Task 2: Calculate cost (abstraction of cost computation).
 Task 3: Compare costs (abstraction of comparison operation).
Method Naming:
 Task 1: distribute_marbles
 Task 2: calculate_cost
 Task 3: compare_costs
Subproblem Interactions:
 We need to distribute marbles before we can calculate the cost (Task 1 before Task 2).
 We must calculate all costs before comparing them (Task 2 before Task 3).
 Task 3 relies on the results of Task 1 and Task 2, so there’s a clear dependency chain: Task 1 > Task 2 > Task 3.
From Brute Force to Optimal Solution
Let’s start by discussing a brute force solution to the problem.
Brute Force Solution:
In the context of this problem, the brute force solution would involve generating all possible ways to distribute the marbles into bags and then calculating the cost for each distribution to identify the minimum and maximum scores.
This would involve the following steps:
Generate all possible combinations of distributing marbles into bags. This could be done using recursion or backtracking.
For each combination, calculate the cost of the distribution by summing the weights of the marbles in each bag according to the rules provided.
Track the minimum and maximum costs observed across all distributions.
Inefficiencies of the Brute Force Solution:
The brute force solution is computationally expensive and becomes infeasible with larger inputs because it generates and checks every possible distribution of marbles into bags.
This means the time complexity of the brute force solution is O(n!), where n is the number of marbles, as in the worst case we might have to generate all permutations of the marbles.
The space complexity is also high due to the storage of all possible distributions, making this approach impractical for large datasets.
Optimization Process:
Instead of generating all possible distributions, we can follow a greedy approach to distribute the marbles.
Sum of adjacent weights: Instead of considering individual weights, we can combine the weights of adjacent marbles and consider these sums for distribution. This strategy aligns with the problem’s cost function which involves the sum of weights of two adjacent marbles.
Sorting: Once we’ve calculated the sums of adjacent weights, we sort them. This is because the problem involves minimizing and maximizing scores and sorting gives us a direct way to access the smallest and largest sums.
Selecting distributions: Once the sums are sorted, we select the first k1 sums for the minimum score and the last k1 sums for the maximum score. The difference between these scores would be our answer.
Impact on Time and Space Complexity:
The optimized solution significantly improves the time and space complexity.
Calculating the sum of adjacent weights and sorting them takes O(n log n) time, where n is the number of marbles. Selecting the minimum and maximum scores then takes O(k) time, where k is the number of bags. Therefore, the overall time complexity of the optimized solution is O(n log n + k).
The space complexity is also improved to O(n), as we’re storing the sums of adjacent weights in a list and not generating all possible distributions.
In conclusion, this optimized approach uses a combination of greedy strategy and sorting to significantly improve upon the brute force solution, making it more efficient and practical for larger inputs.
Code Explanation and Design Decisions
Identify the initial parameters and explain their significance in the context of the problem statement or the solution domain.
Discuss the primary loop or iteration over the input data. What does each iteration represent in terms of the problem you’re trying to solve? How does the iteration advance or contribute to the solution?
If there are conditions or branches within the loop, what do these conditions signify? Explain the logical reasoning behind the branching in the context of the problem’s constraints or requirements.
If there are updates or modifications to parameters within the loop, clarify why these changes are necessary. How do these modifications reflect changes in the state of the solution or the constraints of the problem?
Describe any invariant that’s maintained throughout the code, and explain how it helps meet the problem’s constraints or objectives.
Discuss the significance of the final output in relation to the problem statement or solution domain. What does it represent and how does it satisfy the problem’s requirements?
Remember, the focus here is not to explain what the code does on a syntactic level, but to communicate the intent and rationale behind the code in the context of the problem being solved.
Coding Constructs
 Highlevel problemsolving strategies or techniques being used by this code:
This code uses the strategy of sorting and array manipulation to solve the problem. Specifically, it uses a greedy algorithm which involves sorting an array and then processing the smallest and largest elements to compute the required result. It sorts an array of sums of adjacent pairs in increasing order, and then finds the difference between the sum of the largest ‘k1’ and smallest ‘k1’ elements.
 Explaining the purpose of this code to a nonprogrammer:
Imagine you have a line of marbles and you want to divide these marbles into ‘k’ groups such that each group contains consecutive marbles. Now, the ‘score’ of each group is calculated by summing the first and last marble in the group. The goal of this program is to calculate the maximum possible difference between the highest scoring group and the lowest scoring group.
 Logical elements or constructs used in this code, independent of any programming language:
The logical constructs used in this code are array/list creation, iteration (using a for loop), sorting, arithmetic operations (sum and difference), variable assignments, and function creation and invocation.
 Algorithmic approach used by this code in plain English:
The program begins by creating a list that contains the sum of every two adjacent marbles. It then sorts this list in increasing order. The program calculates the difference between the sum of the largest ‘k1’ pairs and the smallest ‘k1’ pairs. This difference is the maximum possible difference in scores that we can get by dividing the marbles into ‘k’ groups.
 Key steps or operations this code is performing on the input data, and why:
 Step 1: Creating a list of the sum of every two adjacent marbles: This is because the score of a group depends on the sum of the first and last marble, which are adjacent in the sorted array.
 Step 2: Sorting this list: This allows us to easily access the smallest and largest sums.
 Step 3: Calculating the difference between the sum of the largest ‘k1’ pairs and the smallest ‘k1’ pairs: This is the main logic of the problem as per the problem statement.
 Algorithmic patterns or strategies used by this code, irrespective of the specific programming language syntax:
The main algorithmic pattern used here is a greedy algorithm, where we try to get the maximum and minimum sums for the groups to maximize the difference between them. Other techniques used include array manipulation and sorting. The strategy is to sort an array in increasing order and perform mathematical operations on the smallest and largest elements.
Language Agnostic Coding Drills
Dissect the code and identify each distinct concept it contains:
Variable Assignment: Assigning a value to a variable is the most basic concept in coding. In this code, it is used to initialize the variables
n
,pair_weights
, andanswer
.Array/List Manipulation: The ability to create and manipulate arrays or lists is another basic concept. Here, it’s used to create the array
pair_weights
, to store the sum of every two adjacent marbles, and to access these values later.Looping: Looping through elements in a sequence is a common task in programming. In this case, a loop is used to calculate the sum of adjacent marbles and to calculate the sum of the largest and smallest ‘k1’ values in the sorted array.
Conditional Statements: While there’s no explicit
if
condition in the provided code, the code implicitly includes a condition check inside the forloop to terminate when it has processed ‘k1’ elements.Sorting: Sorting is used to organize data in a certain order (ascending or descending). Here, sorting is used to organize the sums of the weights of adjacent marbles in ascending order.
Arithmetic Operations: Arithmetic operations include addition, subtraction, multiplication, and division. In this code, addition is used to calculate the sum of adjacent marbles, and subtraction is used to calculate the difference between the sums.
Function Definition: A function is a block of code designed to perform a particular task. This code is encapsulated within a function named
putMarbles
.
Increasing difficulty order:
Variable Assignment: It’s the most basic concept and is the foundation of any coding task. Without being able to assign variables, we would not be able to store and manipulate data.
Arithmetic Operations: These are basic operations that are essential for any calculation in programming. They’re slightly more complex than variable assignment as they require understanding of mathematical operations.
Array/List Manipulation: Requires understanding how data is stored and accessed in an array or list, and is thus more complex than basic variable assignment and arithmetic operations.
Looping: Iteration or looping requires understanding how to perform repeated operations on a set of data, making it more complex than simple array manipulation.
Sorting: While builtin functions make sorting simple to implement, understanding how sorting algorithms work and their time complexity can make this concept more complex than simply looping over a data structure.
Function Definition: Writing a function requires understanding of how to encapsulate a series of steps into a reusable block of code, which could be considered more complex as it involves understanding the flow of a program and structuring code efficiently.
Problemsolving approach:
The problemsolving approach for this problem involves understanding the problem, breaking it down into simpler parts, and using the identified concepts to create a solution.
Understanding the Problem: The problem involves calculating the difference in score between the highest and lowest scoring groups of marbles. This requires a clear understanding of the given conditions and desired output.
Breaking Down the Problem: We first break down the problem into two main parts: calculating the score of each group of marbles and finding the difference between the highest and lowest scoring groups.
Variable Assignment: We start by assigning the necessary variables, which includes the length of the list of weights and initializing an empty list for the pair weights.
Array Manipulation and Looping: We then calculate the sum of adjacent marbles by using a loop to iterate through the weights list and store the sum in the pair_weights list.
Sorting: We sort the pair_weights list in order to easily access the smallest and largest values.
Arithmetic Operations and Looping: We use another loop to calculate the sum of the largest ‘k1’ and smallest ‘k1’ pair sums, and find their difference.
Function Definition: We encapsulate the above steps in a function to provide a clean, modular solution to the problem.
By practicing these individual coding drills, one could increase their coding proficiency and become better equipped to solve similar problems in the future.
Targeted Drills in Python
Dissect the code and identify each distinct concept it contains:
Variable Assignment: Assigning a value to a variable is the most basic concept in coding. In this code, it is used to initialize the variables
n
,pair_weights
, andanswer
.Array/List Manipulation: The ability to create and manipulate arrays or lists is another basic concept. Here, it’s used to create the array
pair_weights
, to store the sum of every two adjacent marbles, and to access these values later.Looping: Looping through elements in a sequence is a common task in programming. In this case, a loop is used to calculate the sum of adjacent marbles and to calculate the sum of the largest and smallest ‘k1’ values in the sorted array.
Conditional Statements: While there’s no explicit
if
condition in the provided code, the code implicitly includes a condition check inside the forloop to terminate when it has processed ‘k1’ elements.Sorting: Sorting is used to organize data in a certain order (ascending or descending). Here, sorting is used to organize the sums of the weights of adjacent marbles in ascending order.
Arithmetic Operations: Arithmetic operations include addition, subtraction, multiplication, and division. In this code, addition is used to calculate the sum of adjacent marbles, and subtraction is used to calculate the difference between the sums.
Function Definition: A function is a block of code designed to perform a particular task. This code is encapsulated within a function named
putMarbles
.
Increasing difficulty order:
Variable Assignment: It’s the most basic concept and is the foundation of any coding task. Without being able to assign variables, we would not be able to store and manipulate data.
Arithmetic Operations: These are basic operations that are essential for any calculation in programming. They’re slightly more complex than variable assignment as they require understanding of mathematical operations.
Array/List Manipulation: Requires understanding how data is stored and accessed in an array or list, and is thus more complex than basic variable assignment and arithmetic operations.
Looping: Iteration or looping requires understanding how to perform repeated operations on a set of data, making it more complex than simple array manipulation.
Sorting: While builtin functions make sorting simple to implement, understanding how sorting algorithms work and their time complexity can make this concept more complex than simply looping over a data structure.
Function Definition: Writing a function requires understanding of how to encapsulate a series of steps into a reusable block of code, which could be considered more complex as it involves understanding the flow of a program and structuring code efficiently.
Problemsolving approach:
The problemsolving approach for this problem involves understanding the problem, breaking it down into simpler parts, and using the identified concepts to create a solution.
Understanding the Problem: The problem involves calculating the difference in score between the highest and lowest scoring groups of marbles. This requires a clear understanding of the given conditions and desired output.
Breaking Down the Problem: We first break down the problem into two main parts: calculating the score of each group of marbles and finding the difference between the highest and lowest scoring groups.
Variable Assignment: We start by assigning the necessary variables, which includes the length of the list of weights and initializing an empty list for the pair weights.
Array Manipulation and Looping: We then calculate the sum of adjacent marbles by using a loop to iterate through the weights list and store the sum in the pair_weights list.
Sorting: We sort the pair_weights list in order to easily access the smallest and largest values.
Arithmetic Operations and Looping: We use another loop to calculate the sum of the largest ‘k1’ and smallest ‘k1’ pair sums, and find their difference.
Function Definition: We encapsulate the above steps in a function to provide a clean, modular solution to the problem.
By practicing these individual coding drills, one could increase their coding proficiency and become better equipped to solve similar problems in the future.
Q&A
It look like DP, but literally it’s Greedy.
In literal sense you need to make K+1 partitions to find K values including 0th partition and nth partition , but since the first and last index of weights will always be part of both minimum and maximum value so so 2 partitions(0th and nth) are gone, now we are left with k1 partitions to find.
Yes, exactly. Your understanding is correct.
In a sequence of marbles, we can form k subarrays (or partitions), but the first and last marble (or subarray’s ends) will always be part of both minimum and maximum possible scores. So, effectively, we are left with k1 positions to partition the sequence.
The rest of the algorithm is to find the sum of the k1 smallest and largest adjacent pairs and calculate their difference. This is because for any subarray, the “cost” of a subarray in terms of score is determined by the sum of the weights of its ends, which in this case are the two marbles that make up the partition points. The minimum possible score uses the smallest k1 pair sums, while the maximum possible score uses the largest k1 pair sums.
Your explanation helps clarify the logic of the algorithm further.
[1 3 5 1] 13 5 1 > [1]+1 + 3+[1] <— min 1 35 1 > [1]+3 + 5+[1] <— max 1 3 51 > [1]+5 + 1+[1] diff will cancel [1,1] and [1,1] fixed pairs and leave only (3+5)  (1+3) = 84
Yes, you’re right. The given array is split into two parts (subarrays) to get the minimum and maximum sums. Here, the ‘’ denotes the split.
For the minimum sum, you’re splitting the array after the first element [1] and the remaining part [3,5,1]. You add the first and last elements of both parts, which gives you [1]+1 + 3+[1] = 6.
For the maximum sum, you’re splitting the array between the second and third elements [1,3] and [5,1]. You add the first and last elements of both parts, which gives you [1]+3 + 5+[1] = 10.
The difference between the maximum and minimum sums is indeed 10  6 = 4, which matches your explanation.
This demonstrates the reasoning behind the algorithm, namely splitting the array in different ways to find the maximum and minimum sums and then taking the difference between those sums.
The logic in this algorithm depends on the sorted pair sums of adjacent marbles (weights), and the chosen splits (k1 of them) represent the places where we extract the maximum and minimum possible scores.
Also you can calculate sum of pairs in the input vector to reduce space complexity to O(1)O(1)O(1). Even more, you don’t need to sort the whole vector of pairs, you can use quickselect algorithm to place the kth and (n  1  k)th elements on their places and them just sum every element before them. This improves time complexity to linear (O(n)O(n)O(n)) in average.
That’s a very insightful observation, which shows your understanding of more advanced concepts related to algorithmic complexity.
To explain to others:
Space Complexity: By calculating the sum of pairs in the input vector itself, we avoid the need to create an additional list, which reduces the space complexity from O(n) to O(1). In simpler terms, we’re using less memory.
Time Complexity: The quickselect algorithm is an efficient inplace variation of the quicksort algorithm. It allows us to select the kth smallest (or largest) element from an unsorted list in linear time on average, as opposed to sorting the entire list, which can take quadratic time in the worst case. This can significantly improve the performance of the algorithm, especially with large inputs.
However, note that the quickselect algorithm has a worstcase time complexity of O(n^2), which occurs when the pivot element is the smallest or the largest element in the list. This can be mitigated by using a ‘median of medians’ approach to choose the pivot, which ensures a worstcase time complexity of O(n). However, this approach has a larger constant factor and is more complex to implement, so it’s generally only used when worstcase performance is a concern.
These kinds of improvements show the importance of understanding both the problem and the algorithms available to you, as well as how to analyze the time and space complexity of your solutions. Always remember to consider both the practical and theoretical aspects of your solution.
To be honest I thought it is a binary search question something like max allocation of pages in a book.
There are some similarities between this problem and the “Allocation of Maximum Number of Pages” problem. Both involve dividing elements (books or marbles in our case) into a certain number of groups.
However, the difference lies in the constraints and objective of each problem:
In the “Allocation of Maximum Number of Pages” problem, the aim is to minimize the maximum sum of pages assigned to a student (where each student is assigned contiguous books). This objective necessitates binary search for an optimal allocation.
In our “Marble Allocation” problem, we’re not trying to minimize the maximum sum within a partition or constrain the sum in any way. Instead, we’re trying to maximize and minimize the sum of the first and last elements in each partition. The process involves sorting pairs of contiguous elements and then selecting specific sums based on their positions in the sorted array.
So while both problems involve partitioning an array, the rules and objectives of the partitioning are different, leading to different solution strategies. The “Marble Allocation” problem can be solved more directly by sorting and summing, without the need for binary search.
However, your thought process is good! Trying to apply known strategies to new problems is a key part of problemsolving in computer science. Sometimes a problem can be solved with multiple approaches. It’s always good to consider different strategies and choose the most efficient one based on the problem’s constraints and requirements.
If the length of the input is 10^5 or greater, you need a solution better than O(N^2) to avoid TLE.
Indeed, an O(N^2) time complexity would be too slow for an input size of 10^5. It is crucial to come up with an algorithm with a lower time complexity to avoid a Time Limit Exceeded (TLE) error for such large inputs.
The given solution has a time complexity of O(N log N) due to the sorting of the pair_weights array. In practice, an O(N log N) algorithm would be able to handle an input size of 10^5 comfortably, because log(10^5) is a reasonably small number (around 17 if base 2).
However, if we want to further optimize the algorithm, we can use a selection algorithm (such as the QuickSelect mentioned in a previous comment) to find the k1 smallest and largest elements, instead of sorting the entire array. This would bring down the average case time complexity to O(N), making it even more efficient for large inputs.
QuickSelect has a worstcase time complexity of O(N^2), but this worstcase scenario is unlikely in practice if we use a good pivot selection strategy (such as ‘median of medians’). With a good pivot selection strategy, the time complexity in practice would be close to linear for most inputs.
It took a moment to figure out what happens with the first and last weights. Turns out that these are eliminated when we subtract the min pairWeight from the max, e.g. with weights = [1, 3, 5, 1], and k = 2
partition with min pairWeight (4): [1][3, 5, 1] => sum = (1 + 1) + (3 + 1) partition with max pairWeight (8): [1, 3][5, 1] => sum = (1 + 3) + (5 + 1) then max pairWeight  min pairWeight = (1 + 3 + 5 + 1)  (1 + 1 + 3 + 1) = 8  4 = 4
That’s a great breakdown of the algorithm using a simple example! Your explanation captures the essence of the algorithm correctly.
Here is the stepbystep process of what the code does based on your example:
 The list of weights is [1, 3, 5, 1] and we want to split this into k = 2 groups.
 The algorithm computes the pairwise weights for the list [1+3, 3+5, 5+1] = [4, 8, 6] and sorts them to get [4, 6, 8].
 The algorithm then calculates the sum of the smallest k  1 = 1 pairs which is 4 (smallest pair) and the sum of the largest k  1 = 1 pairs which is 8 (largest pair).
 The result is the difference between the sum of the largest pairs and the smallest pairs which is 8  4 = 4.
Your observation is also correct, the weights of the first and the last marble in the list do not influence the outcome because they are included in both the max and min pairWeight calculations, so they effectively cancel each other out in the final difference.
With two heaps we can achieve O(n⋅logK)
That’s correct! A priority queue (or a heap) can be used to improve the time complexity of the algorithm.
Here’s a broad idea of how you could do it:
 Use two heaps, a minheap and a maxheap.
 Iterate through the list of weights and for each pair of weights, add the sum to both heaps.
 Pop off the smallest and largest elements from the minheap and maxheap respectively, for a total of
k1
times.  The sum of the elements in the maxheap now represents the maximum score and the sum of the elements in the minheap represents the minimum score.
 The difference between the maximum score and the minimum score is the result.
The time complexity of this approach is O(n log k)
since you’re doing n
additions to both heaps and k1
deletions from both heaps. Each addition and deletion operation in a heap can be done in O(log k)
time.
This approach uses the property of a heap that allows efficient access to the minimum or maximum element and is faster for larger inputs compared to sorting the whole list of pair weights which takes O(n log n)
time.


why we are taking k  1 min and max ?
The reason we take “k1” minimum and maximum elements is because of the way the problem is framed. We want to split the weights into “k” consecutive groups. In the original code, each group is characterized by the sum of the weights of two consecutive marbles, except for the first and the last group which only contain one marble each.
So if we have “n” marbles and we want to form “k” groups, there will be “k1” “boundaries” or “divisions” between these groups.
Each boundary is defined by two consecutive marbles, hence when we are calculating the sum of the weights for the minimum and maximum score, we only need to consider “k1” pairs, i.e., the “k1” smallest and “k1” largest sums of weights of two consecutive marbles.
That’s why we take “k1” minimum and maximum sums of weights of two consecutive marbles for the calculation of the minimum and maximum score respectively.
Let’s illustrate this with a simple array of weights and a chosen ‘k’.
Let’s take weights = [3, 1, 4, 2, 5]
and k = 3
.
Step 1: Create pair weights and sort them
We create an array with sums of each pair of adjacent marbles:
pair_weights = [3+1, 1+4, 4+2, 2+5] = [4, 5, 6, 7]
, then we sort it:
pair_weights_sorted = [4, 5, 6, 7]
Step 2: Find the minimum and maximum groups
In the sorted array, the first k1
elements represent the smallest sums, and the last k1
elements represent the largest sums.
If we divide weights
array to create k
groups with smallest sums, we’ll get the groups as follows:
 Group 1:
[3, 1]
with sum = 4  Group 2:
[4, 2]
with sum = 6  Group 3:
[5]
with sum = 5
And for largest sums:
 Group 1:
[3]
with sum = 3  Group 2:
[1, 4]
with sum = 5  Group 3:
[2, 5]
with sum = 7
Step 3: Calculate the minimum and maximum score
The minimum score is the sum of the first k1
elements from pair_weights_sorted
(4+5) = 9
The maximum score is the sum of the last k1
elements from pair_weights_sorted
(6+7) = 13
Step 4: Return the difference The difference between the maximum and minimum score is (13  9) = 4.
Visualising this process, the array looks like this:
Original:
weights = [3, 1, 4, 2, 5]
After making pairs:
pair_weights = [4, 5, 6, 7]
Smallest sum groups:
Group 1: [3, 1]
Group 2: [4, 2]
Group 3: [5]
Largest sum groups:
Group 1: [3]
Group 2: [1, 4]
Group 3: [2, 5]
Notice how we have k1
pairs or boundaries between k
groups. Hence, we consider k1
minimum and maximum elements.
Always check the constraints 1 <= k <= weights.length <= 10^5 Your solution is O(n∗k) which is 10^10 operations in worst case. But in all these leetcode problems you need to find an algo that is lower than 10^8. So in this case you must find a solution with O(n∗logn) time complexity, which is sorting.
Absolutely, understanding and taking into account the constraints of the problem is crucial when developing a solution. This helps us avoid TLE (Time Limit Exceeded) errors and ensures our solution is efficient.
In the case of this problem, the constraints are indeed such that an O(nk) algorithm would potentially result in too many operations for larger inputs, leading to an inefficient solution. An algorithm with a time complexity of O(nlog(n)) or better is more desirable, as it scales much better with larger inputs.
Sorting is often used in problems like these to reduce the time complexity. The use of sorting allows us to arrange our data in a way that makes it easier to search for certain elements or sums, and operations like finding minimum or maximum values become much more efficient.
Other techniques such as binary search, heap data structures, or divideandconquer strategies can also be used to meet the constraints and optimize the solution. The use of these techniques often depends on the specifics of the problem, so a good understanding of the problem and various algorithmic strategies is important.
Quite the same as the fundamental method in Combinatorics, the ‘Stars and Bars’ method.
The ‘Stars and Bars’ method is a combinatorics technique used to solve problems of distributing objects into bins or groups. While it doesn’t have a direct application in this problem, it’s worth understanding as a valuable tool in your problemsolving arsenal.
In ‘Stars and Bars’, we consider ‘stars’ as objects to be distributed, and ‘bars’ as dividers that segment these objects into groups. This visual representation can help us derive formulas for counting the number of ways to distribute objects.
Although this problem doesn’t explicitly involve distributing objects into groups, the idea of partitioning an array into segments is somewhat analogous to the ‘Stars and Bars’ method. The weights can be seen as objects (stars), and the splitting points as dividers (bars).
However, this problem takes it a step further by introducing a “score” based on the sum of the ends of each partition, and seeks to find the maximum and minimum possible score. Hence, while the basic partitioning concept is similar, the optimization process is unique to this problem.
Why did it start at ——–>pairWeights[n  2  i] rather than at—————> pairWeights[n  1  i]?
Let’s recall that we’re trying to find the sum of the k  1 largest pairs and the k  1 smallest pairs from the sorted ‘pair_weights’ list. We have to subtract these sums to get our answer.
Now, if we look at the indexes of a list in Python, they start from 0. So, the index of the last element in the list would be ’n  1’, where ’n’ is the total number of elements in the list. In this case, ’n  1’ would be the index of the last element in the ‘pair_weights’ list.
However, we’re trying to get the sum of the k  1 largest elements, not k elements. Therefore, we need to start one position before the last element, which would be at the index ’n  2’. Hence, we start at ‘pair_weights[n  2  i]’. The ‘i’ is used to move backwards from this starting position, allowing us to add up the k  1 largest elements.
Starting at ‘pair_weights[n  1  i]’ would have us sum up k elements, not k  1, which is not what we’re looking for in this problem.
Try to transform this question to the form what I can understand! What is the essential meaning for a partition? Find k1 pairs (weights[P[i]], weights[P[i]+1]) where P[i] denotes the end index for the ith subarray for this partition P . The score for the this partition P is score(P)=weights[0]+weights[n−1]+∑i(weights[P[i]]+weights[P[i]+1])score(P)=weights[0]+weights[n1]+\sum_i (weights[P[i]]+weights[P[i]+1])score(P)=weights[0]+weights[n−1]+∑ i (weights[P[i]]+weights[P[i]+1]) The wanted answer is maxPscore(P)−minPscore(P)\max_P score(P)\min_P score(P)max P score(P)−min P score(P)
Yes, your understanding is correct. The goal of the problem is to find the maximum and minimum possible scores when partitioning the weights into k partitions and calculating the difference between the two.
To understand this more intuitively, let’s think about what a partition really is. When we partition the array into k groups, each group is essentially a subarray of the original array. Each subarray (or partition) will start at one point in the array and end at another, inclusive. Thus, the end points of each subarray (or partition) are crucial in calculating the score.
The score for each partition is the sum of the first weight, the last weight, and the sum of the weights of two adjacent ends (i.e., the sum of a pair (weights[P[i]], weights[P[i]+1])) at each split point.
The problem asks for the difference between the maximum and minimum scores. The maximum score is achieved when the sum of the weights of the k  1 largest pairs is as high as possible, and the minimum score is achieved when the sum of the weights of the k  1 smallest pairs is as low as possible.
To get the answer, we subtract the minimum score from the maximum score:
max(P score)  min(P score)
where P represents a particular partition.
For someone who didn’t understand it mathematically, in the above explanation the idea is to add ends of adjacent subarrays instead of the ends of the same subarray.
Then sort the sums and select the total of k  1 from top and bottom and return the difference.
Exactly! This is a perfect concise summary of the approach.
The crucial observation here is that the “ends” of each partition actually form pairs with the ends of adjacent partitions. In this way, instead of considering each partition independently, we consider the interaction between partitions.
We sum up the weights of the adjacent ends of each partition (i.e., the last weight of a partition and the first weight of the next partition), and store these sums in an array.
Then we sort this array. The maximum possible score is achieved by summing up the largest k  1 pair sums (from the end of the sorted array), and the minimum possible score is achieved by summing up the smallest k  1 pair sums (from the start of the sorted array).
Finally, we return the difference between the maximum and minimum scores. This approach avoids the need to generate all possible partitions of the array, which would be infeasible for large arrays due to the exponential number of possible partitions.
I actually thought this question gonna be solved by DP. Not give a second and started coding build the logic but due to constraint it gives TLE. And also Leetcode sucks at giving hints(in this question):
every bag contain subarray endpoints matter use priority queue
It’s true, it may seem like Dynamic Programming (DP) could be a feasible approach for this problem initially, especially given that it involves partitioning an array and calculating the maximum and minimum scores. However, the constraints indeed rule out a straightforward DP solution due to time complexity, as you’ve noted.
As for the hints provided by LeetCode, it’s worth noting that problemsolving hints are often intentionally vague to encourage a variety of approaches and creative problemsolving. In this case:
“Every bag contains a subarray” is telling you to partition the original array.
“Endpoints matter” is encouraging you to focus on the ends of the subarrays, which is a key point in this problem.
“Use a priority queue” is hinting at a data structure that can efficiently keep track of the smallest or largest elements, which is useful for this problem where we need to find the largest and smallest pair sums.
Even though the hint may not directly suggest the full strategy, it provides a direction to start the problem solving process. This process of interpreting hints and using them to inform your approach is an important part of problem solving in programming competitions and interviews, and is a skill that develops with practice.
why my code not working? I used backtrack to enumerate every possibility. [54,6,34,66,63,52,39,62,46,75,28,65,18,37,18,13,33,69,19,40,13,10,43,61,72] with k = 4 expected 289, but my result is 271. this is really hard to verify by hand…


Your backtracking approach looks conceptually correct for enumerating all possible ways to partition the array, but it does seem to be missing an important part of the problem.
Specifically, in the problem statement, we’re asked to maximize the sum of the first and last marble in each bag, and to minimize the same quantity. Therefore, when you’re creating your partitions, you should add not just the value at the partition index, but also the one next to it.
Also, the way you’re partitioning the array seems to miss the point that we are splitting the array into k+1 subarrays. For instance, if you have the weights [1, 2, 3, 4] and you choose k=1, you’ll have two bags: [1, 2] and [3, 4].
Given the constraints of the problem, a backtracking solution might lead to a Time Limit Exceeded error, since it has a time complexity of O(n^k), which is not feasible for larger inputs. The hints suggest using a priority queue, and a more efficient approach involves sorting a list of pair sums. However, your current approach is a good way to thoroughly understand the problem and how the partitioning works.
Remember that if a bag contains only 1 marble from position ‘i’ then The weight of that bag is weight[ i ] + weight[ i ]
Yes, you’re correct. In the problem statement, it’s mentioned that “The weight of a bag is the sum of the weights of the first and the last marble in the bag.” So if there’s only one marble in the bag, the weight of that marble will be counted twice.
This is a point to keep in mind when analyzing and understanding the problem. When dividing the marbles into bags, it might be beneficial to have more marbles in some bags and only one in others, depending on the distribution of weights. This will also be significant when calculating the maximum and minimum total weights of the bags.
Focus on the boundaries, not the marbles.
The problem description may look like it failed to mention you should use up all the marbles, but this is actually a hint in disguise. Because if you are tricked into thinking you can simply find k bags with each bag having exactly one marble and leaving out the rest of the marbles, then you will come to a “solution” that the k smallest marbles and k largest marbles are the answer. Well, you are wrong but actually quite close to the real one. You just need to see the problem from a new angle and shift the focus from the marbles to the boundaries between the marbles.
That’s a very insightful observation! In this problem, it’s indeed important to think about the “boundaries” between the marbles, not the marbles themselves.
In essence, the problem can be reframed as “how to place k1 ‘boundaries’ between the marbles such that the difference between the maximum and minimum total weight of the bags is minimized”. The first and the last marbles are always the endpoints of some bags and are used in both the maximum and minimum total weights, thus the problem boils down to how we distribute the remaining ‘boundaries’.
If we look at the problem this way, we can start to see why the editorial solution works. Each ‘boundary’ splits a pair of adjacent marbles, adding the sum of their weights to the total weight. The sums of the weights of adjacent marble pairs can be precomputed, sorted, and then the largest k1 and smallest k1 sums are chosen to maximize and minimize the total weights, respectively.
Problem should have condition to use all marbles. what if i don’t want to put marble at 0th index in bag and i goes for another marble. In example 1: min is (1+1) + (1+3) but if i=j is possible then i can take min as (1+1) + (1+1), similarly for maximum i can take (3+3) + (5+5). But clearly it’s not the case.
You’re correct. The problem requires us to use all marbles and partition the marbles into “k” bags. The marbles at the 0th and the last index must always be included in some bag because we’re partitioning the entire array of marbles, and not selecting a subset of the array.
So the scenario you described, where we leave out a marble at the 0th index, doesn’t fit within the constraints of the problem. In the same vein, since we’re required to make “k” bags, choosing the minimum or maximum sum in ways that result in less than “k” bags isn’t allowed either.
The objective is to find the configuration of “k” bags that would yield the maximum and minimum total weights, and then return the difference between those two totals. So in essence, the problem is about dividing the array into “k” partitions such that we maximize the difference between the total weight of the heaviest and lightest possible bag configurations.
This problem does indeed require a careful reading of the problem statement to fully understand the constraints and the goal. It’s a good reminder to always thoroughly read and understand the problem before diving into coding a solution.
This is a nice problem, but it becomes even more interesting if we add another constraint, that is no bag can contain exactly one marble.
Now the convenience that we were enjoying earlier is all gone as, now if we have chosen iii and i+1i+1i+1 indices for MaxScoreMaxScoreMaxScore then we can no longer choose i+1i+1i+1 and i+2i+2i+2 indices for the same. I don’t know the solution yet, but would love to have constructive inputs from the community.
Adding a constraint that no bag can contain exactly one marble does indeed make this problem more complex. It introduces a new dependency in our decisions  if we choose to make a bag using marbles i and i+1, we can’t then create another bag with i+1 and i+2. This is because the marble at index i+1 would already be in a bag.
With this additional constraint, the problem becomes a variant of the Interval Scheduling problem, which is typically solved using dynamic programming or greedy algorithms. The Interval Scheduling problem asks for the maximum number of nonoverlapping intervals (or in this case, bags) that can be selected.
However, there’s a difference in your problem  we’re not trying to maximize the number of bags (it’s fixed to k), but instead the difference between the maximum and minimum total weights of the bags.
A possible approach might be to first sort all pairs of adjacent marbles by their total weights. Then, apply a modified Interval Scheduling algorithm, but instead of always choosing the next available bag with the earliest end, you could try to make a decision based on the potential impact on the maxmin difference.
You’d also need to update your definition of “overlapping”  now two bags overlap if they share a marble, not just if their ranges overlap.
This would certainly be more complex than the original problem, and might require a more sophisticated data structure to track the potential bags and their overlaps efficiently. It would be a fun challenge though!
As always, you should start by clearly defining the problem and any constraints, and then work through some examples by hand to see if you can spot any patterns or insights that might lead to an algorithm.
Think of it this way, you have to partition the array in k different windows. Since you need all elements, the first and last elements will always be present in the sum. Now whenever you place a partition, the two elements adjacent to partition are added to the sum. So how would you maximize or minimize the sum.
Right, to maximize the sum you would place the partition so that it is adjacent to the two highest weights in the array. This is because these highweight marbles would then be included in the sum, making it as large as possible. Conversely, to minimize the sum you would place the partition so that it is adjacent to the two lowest weights. This would contribute the least amount to the sum, thus minimizing it.
To ensure that you pick the highest or lowest pairs of adjacent marbles, you can sort the array of pair weights in decreasing order. For the maximum sum, you would then pick the first k1 pairs from this sorted list. Similarly, for the minimum sum, you would pick the last k1 pairs.
The resulting difference between the maximum and minimum sums would be the maximum possible difference, which is the answer to the problem.
It’s also important to note that we subtract 2 from k (k2) because the first and last marbles are always included in both the minimum and maximum sums, leaving us with k2 marbles to choose for our partitions.
Think about partitions instead of marbles.
That’s an excellent way to approach this problem! When we consider the problem in terms of creating partitions, it becomes a lot clearer.
We are trying to create k partitions of the array. Each partition results in a subarray (or a bag of marbles), and the sum of the end values of each partition contributes to our total sum. The first and last weights of the entire array will always be a part of our sum, because no matter how we partition the array, these values are the endpoints.
Now, when we add a partition, we essentially add two new endpoint values to our total sum. For the maximum sum, we want these new endpoints to be as large as possible. Conversely, for the minimum sum, we want these new endpoints to be as small as possible.
By sorting the array of pair weights (which are essentially our potential new endpoints), we can easily select the k1 largest values for our maximum sum, and the k1 smallest values for our minimum sum.
So, thinking in terms of partitions can indeed make the problem solving process more manageable.
As a beginner can someone pls help me with finding the intution for this, all i can understand are: >We have to do partitions according to some rule. >The first and last element are the elements going in the beginning and last of the bag(s)…
Let’s break it down:
Problem understanding: You’re given a list of weights (marbles), and you’re allowed to make ‘k’ partitions in this list. A partition means dividing the list into segments. Each segment will go into a separate bag. The weight of a bag is calculated as the sum of the weight of the first and last marble in the bag.
Rule for partitions: You need to create the partitions such that you maximize the total weight of all the bags and also minimize the total weight of all the bags.
The challenge: You have to return the difference between the maximum total weight and the minimum total weight.
Here’s the key insight:
Intuition: The first and last marble in the list will always contribute to the total weight (since they will always be the first and last marbles in some bag, no matter how you partition the list). Therefore, we need to think about what happens when we add a partition.
When we create a partition, we essentially add a new bag. The weight of this new bag is the sum of the first and last marble in it, which are the marbles on either side of the partition in the original list.
So to maximize the total weight, we want to put partitions next to the heaviest marbles. And to minimize the total weight, we want to put partitions next to the lightest marbles.
With this intuition, we can create an array of pair weights (each pair is the weight of marbles on either side of a potential partition). Sorting this array, we get the marbles sorted by weight.
Then we take the k1 heaviest pair weights for the maximum total weight (since the first and last marbles in the list are always included, we only need k1 more partitions). Similarly, we take the k1 lightest pair weights for the minimum total weight.
The difference between these two sums gives us our final answer.
I hope this breakdown helps clarify the problem and how to approach it! It’s a tricky one, so don’t worry if it doesn’t click right away.
First and last bag will always have first and last marbles, respectively.
Yes, you’re right. No matter how you partition the array, the first marble will always be the starting point of the first bag, and the last marble will always be the endpoint of the last bag. This is why in our computation, we always add the weights of the first and last marble to both the minimum and maximum total weights. These two marbles will always contribute to the total weight of all bags. The remaining weights will be determined by the specific partitions we make, i.e., where we choose to break the array into different subarrays (or bags).
and cancel out since we are finding the difference
Yes, that’s correct! The weights of the first and last marble are included in both the calculation of the maximum and the minimum score. Thus, when you subtract the minimum score from the maximum score, these weights cancel each other out. That’s why the final answer depends only on the difference between the sums of the weights chosen for partitioning the array into bags. That difference comes from the k1 smallest and k1 largest weights that could be chosen for those partitions.
Imagine we’re finding the sum of the gap between each two bags. ie. Consider (bag[i].right+ bag[i+1].left)
Exactly, this is a very useful perspective for the problem. In the given problem, each time we place a partition, we’re essentially defining the boundary between two bags. The cost of placing this partition, or “the gap” as you’ve referred to it, is determined by the two marbles on either side of this partition.
If we sort the list of all possible gaps, the “k1” smallest gaps would give us the minimum score and the “k1” largest gaps would give us the maximum score.
As the weights of the first and last marble are included in both the maximum and minimum scores, when we calculate the difference between the max and min scores, these values cancel out.
So, the answer to the problem is the difference between the sums of the “k1” smallest and “k1” largest possible gaps.
Hence, the problem is essentially reduced to selecting these gaps, and calculating the difference between their sums.
Is it look Similar to Book Allocation and Painter’s Partition Problem?
While the Book Allocation and Painter’s Partition Problems do share similarities with this problem in that they involve partitioning a given set into subsets, they are not exactly similar.
In the Book Allocation and Painter’s Partition Problems, you’re trying to partition a set into k subsets such that the sum of the values in each subset is minimized or maximized. This often involves binary search and dynamic programming.
In this problem, however, you are partitioning the array into bags (subsets), but the way the “cost” of each subset is determined (by the sum of the first and last elements) is quite different. Also, you are interested in the difference between the maximum and minimum possible total “costs” rather than simply maximizing or minimizing a sum.
So, while the problems are similar in the sense that they involve partitioning a set, the specifics of how this is done and what is being optimized are different.
Once you have realised that DP won’t work, you know there has to be a mathematical trick. At that point, look at the simplest case (in this case k bags) and work out the answer for that (= 0). Then see what happens when you add 1 level of complexity (ie. another bad). What does that to the solution. What happens when we add a second level. Can we work out the rule that gives us the answer with n extra bags?
Yes, this is a great way to approach this problem and many others! Often when faced with a complex problem, breaking it down and starting with the simplest possible case can be very helpful.
In this case, starting with k bags, where each bag contains a single marble, can help you recognize that the “cost” of each bag is the sum of its first and last element  which, when the bag contains a single marble, is just twice the value of that marble. When you add another bag, you start to see how the placement of the partition impacts the total cost  the cost of the bag on the left increases, while the cost of the new bag is the sum of its only marble and the last marble of the previous bag.
Continuing this process, adding more bags one at a time and observing how the total cost changes, you start to see the pattern  that the difference in cost between the maximum and minimum possible total costs is determined by the sums of the first and last elements of the k1 pairs with the highest and lowest sums.
This kind of iterative, stepbystep process is a great way to uncover the underlying patterns and rules in a problem, especially when the problem seems too complex to solve with traditional programming techniques like dynamic programming or depthfirst search. It also helps develop your problemsolving skills and intuition, which will be useful in tackling other complex problems.
First, there are two special cases: when k=1k=1k=1 or n=kn=kn=k, there is just one possible distribution. When k=1k=1k=1 we can do nothing but put all marbles in one bag. When n=kn=kn=k, we can only put one marble in each bag. Since there is just one possible distribution, the difference between the one with maximum score and the one with minimum score would obviously 0.
Now let’s think about the general case. Suppose there are nnn marbles and kkk bags. We can observe that regardless how we put the marbles, the weight of the first marble and the last marble would always be counted. The score differs among different distributions which depends on how we split the marbles in the middle.
Consider the marbles (I will just note the indices below): [0,1,2,…,n−1][0, 1, 2, …, n1][0,1,2,…,n−1]. To split them into kkk different bags, we just need to put k−1k1k−1 bars in the gaps of the marbles. That being said, we have n−1n1n−1 gaps, and we put k−1k1k−1 bars there. When we put a bar after iiith marble, the score of such distribution would increase by weights[i]+weights[i+1]weights[i] + weights[i+1]weights[i]+weights[i+1]
Taking n=10,k=3n=10, k = 3n=10,k=3 as an example, One possible distribution is [0,1,2,∣3,4,5,6,7,∣8,9][0, 1, 2,  3, 4, 5, 6, 7,  8, 9][0,1,2,∣3,4,5,6,7,∣8,9]. When we put a bar after marble 2, the score increases by weights[2]+weights[3]weights[2] + weights[3]weights[2]+weights[3].
Hence, the problem turns into that, with n−1n1n−1 gaps, and the score added after placing a bar after iiith marble is weights[i]+weights[i+1]weights[i] + weights[i+1]weights[i]+weights[i+1], we want to find the minimum/maximum score after placing k−1k1k−1 bars.
With this understanding, the solution is simple: record weights[i]+weights[i+1]weights[i] + weights[i+1]weights[i]+weights[i+1] for each 0 ≤ i < n−10 \leq i < n10 ≤ i < n−1, sort them, select k−1k1k−1 smallest/largest ones from them, and find the difference in score.
Let’s imagine we have a row of marbles, and we need to put them into bags. The trick is that when we put the marbles in the bag, we add up the weight of the first and last marbles in that bag. When we have just one bag or when we have the same number of bags as marbles, there’s only one way we can arrange the marbles, so the difference between the maximum and minimum score is zero  there’s no variation.
But when we have more than one bag and less than one bag per marble, things get interesting. We notice that no matter how we distribute the marbles, the first and last marbles always end up in a bag. The difference in scores between different distributions depends on how we split the marbles in the middle.
Imagine the marbles are laid out in a line. To divide them into bags, we can think of placing dividers or “bars” in between the marbles. Whenever we place a bar, we add the weights of the marbles on either side of the bar to the score.
So, the task is now transformed: with a certain number of gaps between marbles, and knowing that placing a bar in a gap increases the score by the weight of the marbles on either side of the bar, we need to figure out the minimum and maximum score possible when we place a certain number of bars.
To solve this, we record the sum of the weights for each pair of neighboring marbles, sort these sums, then select the smallest and largest sums for the number of bars we need to place, and calculate the difference in score.
It’s a little like figuring out the best places to cut a cake to get the smallest and largest pieces possible.
[3 5 7  8  9] in this partition if we select (7 +8) and (8+9) 8 will be counted twice ?
Yes, you’re correct. In your example, if we partition the marbles such that we have [3, 5, 7  8  9], then the calculation for the score involves adding the weights of the marbles on either side of the partition. Therefore, if we select (7 + 8) and (8 + 9), the weight of the 8 marble will be counted twice. This is part of the approach to this problem  we are trying to select the pairings that will lead to the highest and lowest possible scores, which could involve counting a marble’s weight more than once if it’s positioned between two partitions.
Why get the k1th min and max? Why not the absolute min and absolute max? How do we make the logic jump from ’trying out every single k1 partitions to find the max and min sums’ to ‘just get the k1th max and k1th min of the sum of two consecutive values in the array’?
You might be wondering why we’re looking for the k1 smallest and largest pairs of consecutive marbles’ weights, rather than the absolute smallest and largest weights.
The reason for this is tied to the problem’s constraints and how we’re partitioning the marbles into bags. Let’s try to understand this.
Absolute minimum and maximum values don’t matter: Given the problem’s constraints, each bag must have at least one marble, and we have to distribute all marbles. The absolute minimum and maximum weight values don’t necessarily help us in this situation.
We’re not trying to find the bag with the absolute minimum or maximum weight, but rather the minimum and maximum scores, which are determined by the sum of the weights of the first and last marbles in the bag.
Why k1 pairs? Because we’re partitioning the array into k bags, there will always be k1 pairs of marbles that are at the boundaries between these bags. These pairs contribute to the score because they include the last marble in one bag and the first marble in the next bag.
From all possible partitions to k1 pairs: If we were to try every possible partition, the time complexity would be high and the solution would not be efficient. However, by focusing on the weights of pairs of consecutive marbles, we can find an efficient solution.
The intuition here is that the score depends heavily on these pairs at the boundaries between bags. So, we sum up each pair of consecutive marble weights and sort these sums. The smallest k1 of these sums will help us determine the minimum score and the largest k1 of these sums will help determine the maximum score.
By taking the difference of these two scores, we’re effectively determining the range of scores possible by any partition of the marbles into bags. It’s not about finding the minimum or maximum weight of individual marbles, but about finding the minimum and maximum sum of weights in the configuration that we’re allowed to make.
In summary, the logic jump is made through the understanding of how the partition impacts the calculation of the score and how we can efficiently compute the range of possible scores.
First, there are two special cases: when k=1 or n=k, there is just one possible distribution. When k=1 we can do nothing but put all marbles in one bag. When n=k, we can only put one marble in each bag. Since there is just one possible distribution, the difference between the one with maximum score and the one with minimum score would obviously 0.
Now let’s think about the general case. Suppose there are n marbles and k bags. We can observe that regardless how we put the marbles, the weight of the first marble and the last marble would always be counted. The score differs among different distributions which depends on how we split the marbles in the middle.
Consider the marbles (I will just note the indices below): [0, 1, 2, …, n1]. To split them into k different bags, we just need to put k1 bars in the gaps of the marbles. That being said, we have n1 gaps, and we put k1 bars there. When we put a bar after ith marble, the score of such distribution would increase by weights[i] + weights[i+1]
Taking n=10, k = 3 as an example, One possible distribution is [0, 1, 2,  3, 4, 5, 6, 7,  8, 9]. When we put a bar after marble 2, the score increases by weights[2] + weights[3].
Hence, the problem turns into that, with n1 gaps, and the score added after placing a bar after ith marble is weights[i] + weights[i+1], we want to find the minimum/maximum score after placing k1 bars.
With this understanding, the solution is simple: record weights[i] + weights[i+1] for each 0 <= i < n1, sort them, select k1 smallest/largest ones from them, and find the difference in score.
Here’s a summary:
The problem involves distributing n marbles among k bags with a focus on maximizing and minimizing the score, which is defined by the sum of the weights of the first and last marbles in a bag. When k=1 or n=k, the problem is simple because there’s only one way to distribute the marbles and the difference in scores is zero.
For the general case, you partition the marbles (indexed from 0 to n1) into k bags by placing k1 dividers among them. Each placement increases the total score by the sum of the weights of the marbles adjacent to the divider.
The task then becomes finding the minimum and maximum possible scores by strategically placing the dividers. To do this, you calculate the sum of weights for each adjacent pair of marbles, sort these sums, and select the k1 smallest and largest sums. The difference in scores is then calculated based on these selections.
candidates has length n1. The indices are [0, 1, …, n2]. With i ranges from 0 to n2, n2i gives the correct index backwards.
The phrase “candidates has length n1. The indices are [0, 1, …, n2]. With i ranges from 0 to n2, n2i gives the correct index backwards.” is describing the indexing of a data structure (like an array or a list) named “candidates”.
In many programming languages (such as Python, Java, and C++), the indices of an array or list with length n will start from 0 and end at n1. So for “candidates” with length n1, the indices will start at 0 and end at n2. This is described by “[0, 1, …, n2]”.
Now, when the variable i is looping from 0 to n2 (inclusive), the expression “n2i” will give the correct index to access the elements of “candidates” in a reversed or backwards order.
For example, if i = 0, “n2i” will be “n2” which points to the last element of “candidates”. If i = 1, “n2i” will be “n3”, pointing to the secondtolast element. This continues until when i = n2, “n2i” will be 0, pointing to the first element of “candidates”.
So, this technique is often used when you want to iterate over the elements of an array in reverse order.
Here is a simple Python code snippet illustrating the concept of iterating over an array in reverse order using the formula n2i
:


In this example, the code prints the elements of the candidates
array in reverse order. It starts from the last element (50
), and ends at the first element (10
). This reverse order is achieved by using the index n2i
inside the loop.
Similar Problems
Here are ten problems that have similar concepts. These cover array manipulation, sorting techniques, and exploiting mathematical properties of the problem domain:
215. Kth Largest Element in an Array: This problem is similar because it requires selecting the kth largest element in an unsorted array, similar to our current problem which required identifying the kth smallest and kth largest pairs in an array.
75. Sort Colors: The problem requires partitioning and sorting an array, similar to our current problem where we had to create pairs and sort them.
973. K Closest Points to Origin: In this problem, we have to find K points with the shortest distance to the origin. The underlying concept is similar, because we are looking for the K smallest values based on a particular criterion.
442. Find All Duplicates in an Array: This problem requires an understanding of how to manipulate arrays, similar to the original problem.
287. Find the Duplicate Number: This problem also requires manipulating arrays and identifying unique properties about the elements, much like in our original problem.
56. Merge Intervals: This problem requires sorting and then merging intervals based on specific criteria. The sorting and criteria evaluation can be likened to what was done in the original problem.
4. Median of Two Sorted Arrays: It needs similar skills of dealing with sorted arrays and finding specific statistics (median, in this case) from the array.
152. Maximum Product Subarray: This problem requires you to find the subarray with the maximum product, which is a conceptually similar task to finding the partition with the maximum or minimum sum.
300. Longest Increasing Subsequence: Similar to our problem, this problem also involves understanding the properties of the sequence and exploiting them to find the longest increasing subsequence.
15. 3Sum: The problem requires sorting and then using a twopointer technique to find the three elements that sum to zero. The sorting and use of pointers is somewhat similar to our current problem.
The problems share one or more concepts with our original problem such as handling arrays, dealing with sorted data, identifying and exploiting mathematical properties to derive the solution, partitioning arrays, or using a twopointer technique. Understanding and practicing these problems would help enhance the problemsolving strategies required for solving such problems.