Maximum Distance Between a Pair of Values

class Solution:
    def maxDistance(self, n1: List[int], n2: List[int]) -> int:
        i = j = res = 0
        while i < len(n1) and j < len(n2):
            if n1[i] > n2[j]:
                i += 1
                res = max(res, j - i)
                j += 1
        return res

Identifying Problem Isomorphism

“Maximum Distance Between a Pair of Values” is approximately isomorphic to “Container With Most Water”.


In the “Maximum Distance Between a Pair of Values” problem, we are given two non-increasing arrays, and we are asked to find the maximum distance (j - i) between two valid pairs where i <= j and nums1[i] <= nums2[j].

In the “Container With Most Water” problem, we are given an array where each element represents the height of a line. The task is to find two lines, which, together with the x-axis, forms a container that would hold the most water. Essentially, the task is to maximize the area which can be seen as the product of the height of the shorter line and the distance between the two lines.

The connection between these problems is the idea of finding a pair of elements (or lines) that maximize a certain property (distance in one, area in the other). Both problems also require handling two pointers and utilize similar strategies to move these pointers based on certain conditions.


“Container With Most Water” is a simpler because it only involves a single array. The “Maximum Distance Between a Pair of Values” problem, on the other hand, involves working with two arrays which adds some complexity.

The mapping is approximate due to the fact that the exact property being maximized and the conditions for moving pointers are different between the problems. The difference also lies in how the elements are chosen - in one problem, they come from two different arrays, while in the other problem, they come from the same array.

10 Prerequisite LeetCode Problems

For “1855. Maximum Distance Between a Pair of Values”, the following are a good preparation:

  1. “283. Move Zeroes” - Teaches basic array manipulation which will be beneficial in moving and comparing elements.

  2. “167. Two Sum II - Input array is sorted” - Handling sorted arrays and using two-pointer technique to solve the problem. This approach is often useful in array problems where we have constraints related to indices.

  3. “27. Remove Element” - Another problem about in-place changes in the array, which can help to handle arrays in an efficient manner.

  4. “88. Merge Sorted Array” - Teaches about handling sorted arrays and comparing elements between two arrays.

  5. “350. Intersection of Two Arrays II” - Handling two arrays together and understanding the relationship between their elements.

  6. “334. Increasing Triplet Subsequence” - This problem also deals with the relation between indices and the corresponding elements, which is crucial for the given problem.

  7. “674. Longest Continuous Increasing Subsequence” - Understand how to find longest increasing subsequence in an array, which is beneficial for the concept of distance.

  8. “26. Remove Duplicates from Sorted Array” - Helpful in understanding the manipulation of sorted arrays.

  9. “80. Remove Duplicates from Sorted Array II” - A variation of the previous problem but gives more insights about handling duplicates in a sorted array.

  10. “209. Minimum Size Subarray Sum” - This problem teaches how to manage two pointers and moving them based on certain conditions.

These cover array manipulation, dealing with sorted arrays, and tracking multiple indices simultaneously, which are all relevant to the problem “1855. Maximum Distance Between a Pair of Values”.

Problem Classification

The problem falls under the domain of Array Manipulation and Two Pointers Technique. Specifically, it deals with finding maximum distance pairs across two given arrays under specific constraints.


  • Two non-increasing integer arrays nums1 and nums2.
  • A valid pair of indices (i, j), satisfying certain conditions:
    • (0 \leq i < \text{nums1.length})
    • (0 \leq j < \text{nums2.length})
    • (i \leq j)
    • (nums1[i] \leq nums2[j])
  • The distance of the pair (j - i).
  • The maximum distance among all such valid pairs.
  1. Input: Two non-increasing arrays of integers.
  2. Output: An integer representing the maximum distance between valid pairs.
  3. Constraints: Specific constraints on the valid pair of indices.
  4. Objective: Find and return the maximum distance of any valid pair.

This is an Optimization Problem since we are trying to maximize the distance between a valid pair of indices. It also falls under the Search Problem category because we’re essentially searching for such pairs. Given the need to traverse both arrays to find the pairs, it can also be categorized as a Traversal Problem.

Clarification Questions

  1. Are the input arrays always sorted in a non-increasing order, or do we need to sort them first?
  2. Are duplicate values allowed in the arrays?
  3. What should be returned if there are no valid pairs? Is returning 0 as stated in the problem always acceptable?
  4. Is it guaranteed that the arrays will have at least one element?
  5. Are negative integers or zero allowed in the input arrays, or will they always be positive?
  6. Are there any space or time complexity requirements for the solution?
  7. Is it possible for both arrays to have different lengths?
  8. Do we need to consider multiple maximum distances, or is finding one maximum distance sufficient?
  9. Is the solution expected to be a stand-alone function or integrated into a larger program?
  10. Will the function be called multiple times (thus benefiting from any precomputation), or is it going to be a one-off calculation?

Problem Analysis and Key Insights

  1. Non-Increasing Arrays: Both input arrays are non-increasing, which implies a specific order that can be exploited for efficient searching or matching.

  2. Valid Pair Conditions: The problem has two conditions for a pair to be valid: i <= j and nums1[i] <= nums2[j]. These conditions guide how we can loop through both arrays.

  3. Maximum Distance: The objective is to find the maximum distance j - i for any valid pair. This implies that we need to keep track of a running maximum.

  4. Array Indexes: Both i and j are indexes in the arrays, not the actual values stored at those indexes. This is crucial for calculating the distance.

  5. Constraints: The constraints on array length and element size indicate that the solution needs to be optimized to handle large inputs.

  6. Return 0 if No Valid Pairs: If no valid pairs exist, we need to return 0. This gives us a default value to start with.

  7. Array Lengths: The lengths of the two arrays may be different, which means the solution needs to handle arrays of varying lengths.

  8. Duplication: There is no mention that the values in each array are unique, so the solution must handle duplicate values.

These insights help us understand the problem better and guide us in crafting an optimized solution.

Problem Boundary

The scope of this problem is confined to:

  1. Handling two non-increasing integer arrays, nums1 and nums2.
  2. Identifying valid pairs (i, j) based on given conditions (i <= j and nums1[i] <= nums2[j]).
  3. Calculating the distance j - i for each valid pair.
  4. Finding and returning the maximum distance among all valid pairs.
  5. Returning 0 if no valid pairs exist.

The problem does not extend to:

  1. Sorting or modifying the given arrays.
  2. Handling data types other than integers.
  3. Handling null or empty arrays.

This focus narrows down what we need to consider when coming up with a solution.

Establishing the boundary of this problem involves understanding the limitations and specific conditions outlined in the problem statement:

  1. Array Lengths: Both arrays have a minimum length of 1 and a maximum length of (10^5).

  2. Array Elements: The values in the arrays are integers, ranging from 1 to (10^5).

  3. Array Properties: Both arrays are non-increasing. This constraint narrows down the kind of array manipulations you might consider.

  4. Pair Conditions: A pair ((i, j)) is valid if (i \leq j) and (nums1[i] \leq nums2[j]).

  5. Output: The output is a single integer representing the maximum distance (j - i) among all valid pairs, or 0 if no valid pairs exist.

By adhering to these conditions and limitations, you set the boundary within which the solution must operate.

Distilling the Problem to Its Core Elements

  1. Fundamental Concept: The fundamental concept this problem hinges on is array traversal and comparison. Specifically, you’re comparing elements between two non-increasing arrays to find valid index pairs that satisfy certain conditions.

  2. Simplest Description: Imagine you have two lists of numbers. You’re trying to find pairs of numbers, one from each list, where the second number is greater than or equal to the first. You also want the index of the second number to be greater than or equal to the index of the first. Your goal is to find the pair of numbers that are farthest apart in the lists.

  3. Core Problem: Find the maximum distance (j - i) for all valid pairs ((i, j)) in two given non-increasing arrays.

  4. Key Components:

    • Identifying valid pairs ((i, j)) according to the rules: (i \leq j) and (nums1[i] \leq nums2[j]).
    • Calculating the distance (j - i) for each valid pair.
    • Finding the maximum distance among all valid pairs.
  5. Minimal Set of Operations:

    • Initialize variables to keep track of maximum distance and current indexes (i) and (j).
    • Traverse the first array, (nums1), using index (i).
    • For each (i), find a suitable (j) in (nums2) such that (nums1[i] \leq nums2[j]).
    • Calculate (j - i) and update the maximum distance if this distance is larger.

By breaking the problem down like this, you can systematically address each component to arrive at a solution.

Visual Model of the Problem

Visualizing this problem can be helpful to grasp the underlying mechanics. You can use a few different approaches:

  1. Array Diagrams: Draw two parallel lines of boxes to represent the two arrays, nums1 and nums2. Place the array values in the boxes. Draw arrows from elements in nums1 to corresponding valid elements in nums2 that satisfy the conditions. The length of the arrow could visually represent the distance (j - i).

  2. Graphs: Plot the values of nums1 and nums2 on separate lines on a graph. The X-axis represents the index and the Y-axis represents the value. Use vertical lines to connect valid pairs. The length of these vertical lines will represent the distance (j - i).

  3. Tables: Create a table with nums1 on one axis and nums2 on another axis. Mark cells that represent valid pairs. Each cell’s coordinates correspond to (i) and (j), and you can easily see the distances by counting cells horizontally or vertically.

  4. Number Lines: Place two number lines, one for each array. Mark points representing the elements at their respective indexes. Draw arrows between valid pairs.

Each of these visualizations can give you insights into how to find valid pairs efficiently and how to calculate the maximum distance among these pairs.

Problem Restatement

You have two arrays, nums1 and nums2, both sorted in descending order. Your task is to find the maximum difference in indexes (j - i) for pairs where (i) is from nums1 and (j) is from nums2, under the conditions:

  1. (i \leq j)
  2. nums1[i] ( \leq ) nums2[j]

If no such pairs exist, return 0.

Both arrays will have lengths ranging from 1 to (10^5), and the elements in these arrays are between 1 and (10^5).

Abstract Representation of the Problem

You have two ordered sets, A and B, and each has a sequence of elements. You’re looking to find the maximum distance (d = j - i) between two indices (i) in A and (j) in B, such that:

  1. (i \leq j)
  2. (A[i] \leq B[j])

Return the maximum (d) that satisfies these conditions. If no such pairs exist, return 0.

The focus here is on the relative ordering and positioning of elements in two sets rather than on the specific numerical values of the elements.


  1. Non-Increasing Array: An array where elements are equal to or greater than the elements that follow them. This term is crucial because both input arrays are non-increasing, which influences the algorithm’s logic.

  2. Valid Pair: A pair ( (i, j) ) where ( i ) is from the first array and ( j ) is from the second, and both ( i \leq j ) and ( \text{nums1}[i] \leq \text{nums2}[j] ) hold. Understanding what constitutes a valid pair is essential for solving the problem.

  3. Distance: In this context, it refers to the difference ( j - i ) for any valid pair ( (i, j) ). Maximizing this distance is the core problem.

  4. Constraints: These are limitations on the input size or the values within the arrays. Understanding these helps in determining the feasible solution space and optimizing the algorithm.

Each of these terms shapes the rules and the goal of the problem. Understanding them is key to formulating an effective solution.

Problem Simplification and Explanation

The problem asks you to find two numbers, one from each array, that meet specific conditions. You want to maximize the distance between their positions in the arrays.

Key Concepts:

  1. Array: Think of each array as a list of numbers in descending order.

  2. Pair: Imagine picking one number from each list and making a pair.

  3. Validity: A pair is valid if the number from the first list is smaller or equal to the number from the second list, and the position in the second list is equal to or greater than the position in the first list.

  4. Distance: The goal is to maximize the gap between the positions of these numbers in their respective arrays.


Imagine two lines of people waiting to get onto roller coasters of different thrill levels. The first line has people willing to go on roller coasters up to a certain thrill level. The second line has people who are looking for a minimum thrill level. You need to match people from both lines so that the thrill-seeker from the first line is satisfied with the coaster’s thrill level in the second line. And, among all such pairs, you’re trying to find the pair that has the most distance between them in their respective lines.


You’re “pairing” elements from two lists while adhering to specific “validity” conditions. Once you’ve identified these valid pairs, you’re trying to find the one that maximizes “distance” between the positions.

Understanding these components and their interactions will help you devise a strategy to solve the problem efficiently.


  1. Non-Increasing Order: Both arrays are given in non-increasing order. This pattern means we can perform a single pass through both arrays without having to backtrack, as we know the elements are already sorted in a particular order.

  2. Bounds on Length: The lengths of nums1 and nums2 are between 1 and (10^5). This constraint tells us that an (O(n \log n)) or (O(n)) algorithm would be efficient, but an (O(n^2)) algorithm would likely be too slow.

  3. Element Range: Elements range from 1 to (10^5). Knowing the numerical bounds can sometimes help with using array-based counting techniques, although that might not be directly applicable here.

  4. Validity Conditions: A pair ((i, j)) is valid if (i \leq j) and (nums1[i] \leq nums2[j]). These conditions can help us skip irrelevant pairs more quickly, especially given that the arrays are sorted. We can move through each array in one pass, validating as we go along.

  5. Goal is Distance, Not Values: We are not concerned with the actual numbers but the indices. Focusing on the distance (j-i) allows us to consider only the arrangement and not the values themselves, potentially simplifying the problem.

  6. Multiple Valid Pairs: There can be multiple valid pairs, but we only need the one that gives us the maximum distance. This means once we find a ‘better’ pair (greater distance), we can forget the previous ones.

  7. Zero as a Lower Bound: The problem specifies that if no valid pair is found, the answer is zero. Knowing this can simplify boundary conditions.

By focusing on these specific characteristics, we can work on an algorithmic approach that takes advantage of the sorted nature of the input and the specific conditions to find the maximum distance efficiently.

  1. Array Order: Both arrays are non-increasing. This hints that a one-pass or two-pointer technique could be a key strategy for an efficient solution, eliminating the need for sorting or backtracking.

  2. Array Length: The maximum length of (10^5) for both arrays indicates that a quadratic-time solution would be too slow. We should aim for a linear or linearithmic solution.

  3. Element Ranges: Elements range between 1 and (10^5). Although this might not directly impact the logic, it informs us that we won’t run into problems related to negative numbers or zero when comparing elements.

  4. Validity Conditions: The conditions for a pair to be valid ((i \leq j) and (nums1[i] \leq nums2[j])) allow us to rule out a lot of candidate pairs without much computation, making the problem more tractable.

  5. Output Lower Bound: Knowing that zero is the answer if no valid pairs are found eliminates the need to check for this special case.

Analyzing these constraints guides us toward aiming for a linear-time algorithm that exploits the sorted nature of the arrays and uses the validity conditions for early elimination of candidates.

Case Analysis

Additional Examples

1. Smallest Arrays - Edge Case

  • Input: nums1 = [1], nums2 = [1]
  • Output: 0
  • Analysis: Both arrays have the minimum length (1). The only valid pair is (0, 0) with a distance of 0.

2. Unmatched Elements - Edge Case

  • Input: nums1 = [10, 9], nums2 = [8, 7]
  • Output: 0
  • Analysis: No element in nums1 is smaller or equal to any element in nums2. So, no valid pair exists.

3. Single Valid Pair - Boundary Case

  • Input: nums1 = [10, 9], nums2 = [15, 10]
  • Output: 1
  • Analysis: The only valid pair is (0, 1) with a distance of 1.

4. Equal Elements in One Array - Special Case

  • Input: nums1 = [5, 4], nums2 = [5, 5]
  • Output: 1
  • Analysis: The valid pairs are (0, 0), (0, 1), and (1, 1). Maximum distance is 1.

5. Equal Arrays - Special Case

  • Input: nums1 = [10, 10], nums2 = [10, 10]
  • Output: 1
  • Analysis: Valid pairs are (0, 0), (0, 1), and (1, 1). Maximum distance is 1.

6. All Valid Pairs - General Case

  • Input: nums1 = [10, 5], nums2 = [15, 10]
  • Output: 1
  • Analysis: All pairs are valid, but the maximum distance is still 1, corresponding to pairs (0, 1) and (1, 1).

7. Multiple Maximum Distances - General Case

  • Input: nums1 = [20, 15, 10], nums2 = [30, 25, 20]
  • Output: 2
  • Analysis: There are multiple pairs with the maximum distance: (0, 2), (1, 2), and (1, 3).

Edge Cases

  1. Smallest Arrays: Both arrays contain only one element.
  2. Unmatched Elements: No element in one array is less than or equal to any element in the other array.

These test cases help us understand different facets of the problem, including the importance of the array’s sorted nature, the effect of equal elements, and the constraints of index ordering. A robust solution should handle all these scenarios efficiently.

Visualizing these cases can involve using graphs, tables, or simple notations. Here’s how you can visualize some of them:

1. Smallest Arrays - Edge Case

  • nums1: [1]
  • nums2: [1]
  • Pairs: (0, 0)
-|---| nums1 (i)
-|---| nums2 (j)

2. Unmatched Elements - Edge Case

  • nums1: [10, 9]
  • nums2: [8, 7]
   10   9
-|----|----| nums1 (i)

   8    7
-|----|----| nums2 (j)

3. Single Valid Pair - Boundary Case

  • nums1: [10, 9]
  • nums2: [15, 10]
   10   9
-|----|----| nums1 (i)

   15   10
-|----|----| nums2 (j)
  • Pairs: (0, 1)

4. Equal Elements in One Array - Special Case

  • nums1: [5, 4]
  • nums2: [5, 5]
    5    4
-|----|----| nums1 (i)

    5    5
-|----|----| nums2 (j)
  • Pairs: (0, 0), (0, 1), (1, 1)

5. Equal Arrays - Special Case

  • nums1: [10, 10]
  • nums2: [10, 10]
    10   10
-|----|----| nums1 (i)

    10   10
-|----|----| nums2 (j)
  • Pairs: (0, 0), (0, 1), (1, 1)

These visualizations can help you understand the problem constraints, the possible valid pairs, and their distances better. You can map the indices (i, j) between the two arrays to see the valid pairs clearly.

Analyzing different cases reveals several key insights:

  1. Smallest Arrays: When either array contains only one element, the maximum distance can only be 0, as (i, j) would be (0, 0).

  2. Unmatched Elements: If all elements in nums1 are greater than those in nums2, then there will be no valid pairs, and the maximum distance should return 0.

  3. Single Valid Pair: Even if there’s just one valid pair, it can offer a non-zero maximum distance depending on the indices involved.

  4. Equal Elements in One Array: When there are duplicate elements in one array, they can create multiple valid pairs with different distances. In this case, you don’t have to check pairs that offer less distance if you’ve already checked pairs that provide more distance.

  5. Equal Arrays: If the arrays are equal and contain duplicates, then all pairs would be valid but might not offer the maximum distance.

These insights can guide the approach to solving the problem. For example, the algorithm can be optimized by:

  • Starting from the end of the arrays, given they are sorted in non-increasing order.
  • Skipping duplicate elements that do not offer a better distance.

Understanding these nuances can help in crafting an efficient algorithm that accounts for all edge cases and boundary conditions.

Identification of Applicable Theoretical Concepts

Several mathematical and algorithmic concepts can be applied to make the problem more manageable:

  1. Two-Pointer Technique: Given that both arrays are sorted in non-increasing order, we can start from the end of both arrays and work our way to the beginning, reducing the need to check all possible pairs.

  2. Monotonicity: The non-increasing nature of the arrays allows us to make certain deductions that can reduce the number of pairs we need to check. For example, if nums1[i] <= nums2[j], then nums1[i] <= nums2[k] for all k >= j.

  3. Greedy Algorithm: Since we’re interested in the maximum distance, we can keep updating a variable with the maximum distance as we go through the valid pairs.

  4. Invariance: The conditions for a valid pair (i, j) offer an invariant that can help us reduce the search space. If we find a valid pair at indices (i, j), then all pairs (i, k) where k > j are also valid. We can use this invariant to optimize our search for the maximum distance.

  5. Dynamic Programming: Although it may be overkill for this specific problem, a DP approach could be used to remember previously calculated valid pairs to avoid redundant work.

By applying these concepts, we can optimize the algorithm to find the maximum distance between a pair of values more efficiently.

Simple Explanation

Imagine you have two rows of blocks, each row from a different set. The first row has blocks of varying heights, and so does the second row. Your task is to pick one block from each row such that the block from the first row is not taller than the block from the second row.

Now, you want to find the two blocks that are farthest apart from each other in their respective rows while still following the rule that the block from the first row can’t be taller than the block from the second row.

Think of it like a game where you’re trying to make the longest possible “bridge” between two rows of blocks, but you can only place a bridge between blocks where the first one isn’t taller than the second one. Your goal is to find the longest bridge you can make.

Problem Breakdown and Solution Methodology

Approach to Solving the Problem

  1. Initial Setup: Begin by setting pointers at the start of both arrays, and a variable to keep track of the maximum distance found so far, initialized to zero.

  2. Iterate Through The First Array: Move through the first array one element at a time. For each element, find its “partner” in the second array that meets the conditions: it should be equal or greater than the current element from the first array.

  3. Calculate Distance: Once the partner element is found, calculate the distance between the indices. If this distance is greater than the maximum distance found so far, update the maximum distance.

  4. Move Smartly: The arrays are non-increasing, so use this property to avoid redundant calculations. If you’ve found a partner for an element, you don’t need to go back for the next elements.

  5. Edge Cases: Make sure to handle when no valid pairs can be found. In that case, the maximum distance remains zero.

  6. Final Result: Return the maximum distance found.


Think of it as two parallel train tracks with stations at different intervals. You want to find the longest possible gap between a station on the first track and a station on the second track, but you can only measure from a station on the first track to a station further down on the second track.

Effect of Changing Parameters

  • Increasing Array Size: The bigger the arrays, the more comparisons. But thanks to the non-increasing property, we avoid unnecessary checks, making the algorithm more efficient than brute-force.

  • Value Ranges: A wider value range might require more iteration, but again the non-increasing nature helps keep things efficient.


Let’s take Example 1 from the problem: nums1 = [55,30,5,4,2], nums2 = [100,20,10,10,5]

  • Start with pointers at 55 and 100. Distance is 0 - 0 = 0
  • Move to 30. You find 100 is a valid partner. Distance is 0 - 1 = -1. Not greater than the max distance, so continue.
  • Next is 5. Now, you can’t stay at 100; you have to move to 10. The distance is 4 - 2 = 2.
  • For 4, you stay at 10. Distance is 4 - 3 = 1. Not greater than max distance.
  • For 2, you can move to 5. Distance is 4 - 2 = 2.

The maximum distance is 2, which is the output.

This way, each step contributes to finding the maximum possible distance between the two arrays while adhering to the conditions.

Inference of Problem-Solving Approach from the Problem Statement

  1. Non-Increasing Arrays: Both arrays are non-increasing, which means as we move from left to right, the elements do not increase. This helps us move the pointers intelligently. Once a suitable “partner” is found for an element in nums1 in nums2, we don’t need to start again from the beginning for the next element.

  2. Valid Pair (i, j): The problem defines a valid pair where i <= j and nums1[i] <= nums2[j]. This is crucial as it forms the basis for calculating the distance. It tells us that our pointer in the second array should never move back.

  3. Distance (j - i): The distance of the pair, j - i, is what we aim to maximize. This is our end goal. Knowing that we need to maximize a value informs the kind of comparisons we need to do as we iterate.

  4. Maximum Distance: This is what we’re tasked to find. Knowing that we have to find the maximum of something suggests we need to keep track of the maximum value as we iterate through possibilities.

  5. Constraints: The size and element limitations guide us on how complex our solution can be. Given that the array length is up to (10^5), we should aim for an algorithm that operates in linear time to be efficient.

  6. Return 0: If no valid pairs are found, the maximum distance is zero. This is an edge case we need to handle explicitly.

Each of these keywords or terms informs specific parts of our problem-solving approach. The non-increasing nature tells us how to move our pointers. The definition of a valid pair tells us what to look for. The distance tells us what to calculate, and the maximum distance tells us what to return. Finally, the constraints give us an idea of how efficient our solution needs to be.

  1. Non-Increasing Arrays: A simple table with the array indexes and values can show the non-increasing property visually. You can easily spot that the values do not increase as you move from left to right. You could also plot the arrays on a graph where the y-axis represents the values and the x-axis represents the indexes. The plot would be either flat or declining.

  2. Valid Pair (i, j): A 2D grid can help here. Place nums1 values on the y-axis and nums2 values on the x-axis. Mark an “X” wherever a pair (i, j) satisfies nums1[i] <= nums2[j] and i <= j. This will give you a visual representation of all valid pairs.

  3. Distance (j - i): To visualize the distance, you could use arrows in the 2D grid from step 2. The length of each arrow corresponds to j - i. The longer the arrow, the greater the distance.

  4. Maximum Distance: In your 2D grid or your diagram with arrows, highlight the longest arrow(s). This would represent the maximum distance you are trying to find.

  5. Constraints: These are usually best represented textually rather than visually, but if you want to visualize it, you could use a simple note or annotation in your diagrams to remind you of the constraints.

  6. Return 0: In your 2D grid, if there are no “X” marks, then you’ll return 0. This could be indicated by a completely blank grid or a note.

These diagrams and tables can be a powerful way to understand the properties of the problem you’re dealing with and how they interact. They provide a visual guide that can help you think through the problem more clearly.

How did you infer from the problem statement that this problem can be solved using ?

Simple Explanation of the Proof

I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?

Stepwise Refinement

  1. Stepwise Refinement of Approach

    • Step 1: Initialize pointers for both arrays and a variable to store the maximum distance.
    • Step 2: Iterate through nums1 with one pointer.
    • Step 3: For each value in nums1, iterate through nums2 with another pointer, but only if the condition nums1[i] <= nums2[j] is met.
    • Step 4: Calculate the distance j - i for each valid pair.
    • Step 5: Update the maximum distance if a greater one is found.
    • Step 6: Continue until you have iterated through both arrays.
    • Step 7: Return the maximum distance.
  2. Granular, Actionable Steps

    • Initialize Variables: Set i = 0, j = 0, maxDistance = 0.
    • Loop Through nums1: Use a for or while loop to iterate through nums1.
    • Nested Loop for nums2: Inside the first loop, initiate another loop to iterate through nums2.
    • Check Conditions: Use if statements to check the conditions nums1[i] <= nums2[j] and i <= j.
    • Calculate Distance: If conditions are met, calculate distance j - i.
    • Update Maximum Distance: Use an if statement to update maxDistance if a greater distance is found.
    • Return Value: Once the loops are done, return maxDistance.
  3. Independent Parts

    • Checking conditions for valid pairs can be done independently for each pair (i, j).
    • Updating the maximum distance can be done independently of the other operations.
  4. Repeatable Patterns

    • The process of iterating through the array and checking conditions is a repeatable pattern.
    • The act of updating the maximum distance whenever a greater distance is found is also repeatable.

The problem essentially consists of two nested loops, each going through one of the two arrays. Within these loops, there’s a repetitive pattern of condition checking, distance calculation, and updating the maximum distance.

Solution Approach and Analysis

Step-by-Step Process:

  1. Initialize Pointers and Maximum Distance: Start with pointers i and j at 0. Initialize maxDistance as 0.

  2. Iterate Through nums1: Use a loop to go through the elements of nums1. The loop will continue until i reaches the end of nums1.

  3. Nested Iteration Through nums2: For each value at i in nums1, start another loop to go through nums2 using j.

  4. Check Pair Validity: Inside the nested loop, only proceed if nums1[i] <= nums2[j] and i <= j.

  5. Calculate and Update Distance: If the pair (i, j) is valid, calculate the distance as j - i. If this distance is greater than maxDistance, update maxDistance.

  6. Termination and Return: Once all pairs have been considered, return maxDistance.


Imagine two lines of students in a classroom. The first line represents nums1 and the second line represents nums2. Each student has a number tag. A student from the first line (nums1) can only high-five a student in the second line (nums2) if their number is less than or equal to the number of the student in the second line, and they are in the same position or ahead in line. The distance between the students who can high-five is what we are trying to maximize.

Effects of Parameter Changes:

  • Length of Arrays: If either array is longer, more iterations are needed, increasing the time complexity.
  • Element Values: The larger the gap between numbers in nums1 and nums2, the more likely it is to find valid pairs and thus a higher maxDistance.

Example Case:

Let’s consider nums1 = [30, 29, 19, 5] and nums2 = [25, 25, 25, 25, 25].

  • Step 1: Initialize i = 0, j = 0, maxDistance = 0.

  • Step 2: Start looping i from 0 to 3 (nums1.length - 1).

  • Step 3: When i = 2 (nums1[2] = 19), start looping j.

  • Step 4: We find nums1[2] <= nums2[2], nums1[2] <= nums2[3], and nums1[2] <= nums2[4].

  • Step 5: Calculate distances: 2 - 2 = 0, 3 - 2 = 1, 4 - 2 = 2. Update maxDistance = 2.

  • Step 6: No more valid pairs with greater distance found. Return maxDistance = 2.

The approach walks you through the two arrays, carefully checking for valid pairs and keeping track of the maximum distance for such pairs.

Identify Invariant

In this problem, the invariant is the condition that defines a valid pair (i, j): both i <= j and nums1[i] <= nums2[j]. These conditions must hold true for any pair to be considered valid throughout the problem. The algorithm’s logic must respect this invariant to find the maximum distance between such valid pairs.

Identify Loop Invariant

What is the loop invariant in this problem?

Is invariant and loop invariant the same for this problem?

Thought Process

The problem asks us to find the maximum distance between valid pairs (i, j) in two given arrays, nums1 and nums2. A pair is valid if i <= j and nums1[i] <= nums2[j]. The distance is j - i.

Basic Thought Process

  1. Understand Constraints: Both arrays are non-increasing. This can guide us to use a two-pointer technique to compare elements efficiently.

  2. Initialize Pointers: Start with two pointers i and j at the beginning of nums1 and nums2 respectively.

  3. Scan and Compare: Move i and j such that the pair (i, j) is always valid according to the conditions. Update the maximum distance if a pair with a greater distance is found.

  4. Optimization: Due to the non-increasing nature of the arrays, if nums1[i] <= nums2[j], we can safely move j to find a pair with a larger distance, as nums1[i] will also be <= all subsequent nums2[k] for k > j.

  5. Max Distance: Keep track of the maximum distance while performing the above operations.

Python Code

def maxDistance(nums1, nums2):
    i = 0
    j = 0
    max_distance = 0

    while i < len(nums1) and j < len(nums2):
        # Check validity of the pair
        if nums1[i] > nums2[j]:
            i += 1
            # Update maximum distance
            max_distance = max(max_distance, j - i)
            j += 1
    return max_distance

Cues in Problem Statement

  1. “non-increasing 0-indexed integer arrays” hints that the elements are sorted in a specific way, suggesting two-pointer techniques could be efficient.

  2. “maximum distance of any valid pair” implies that we need to keep track of the best (maximum) result as we scan through possible pairs.


  1. The non-increasing nature of both arrays offers an optimized way to find valid pairs without scanning all possible combinations.

  2. We only need to move one pointer at a time, either i or j, because if nums1[i] <= nums2[j], then nums1[i] will also be <= all nums2[k] for k > j.

  3. The problem is essentially asking for the maximum gap between indices of two arrays, under certain conditions, which simplifies our task to tracking indices rather than elements.

By following these steps and understanding the cues and constraints, we can solve the problem in a time-efficient manner.

Establishing Preconditions and Postconditions

  1. Parameters:

    • Inputs: Two non-increasing integer arrays, nums1 and nums2.
    • Types: Both are lists of integers.
    • Representation: nums1 and nums2 represent the arrays in which we need to find valid pairs (i, j) that satisfy given conditions.
  2. Preconditions:

    • State: No specific state of the program is required before this method is called.
    • Constraints: The lengths of nums1 and nums2 should be between 1 and 10^5, and each element should be between 1 and 10^5. Both arrays should be non-increasing.
    • Specific State: None.
  3. Method Functionality:

    • Expected Action: The method finds the maximum distance between valid pairs (i, j) as per the conditions in the problem statement.
    • Interaction: It reads the input arrays and iteratively compares elements to find valid pairs, updating the maximum distance found so far.
  4. Postconditions:

    • State: The method returns the maximum distance between valid pairs, and no state is altered.
    • Return Value: An integer representing the maximum distance.
    • Side Effects: None.
  5. Error Handling:

    • Response: The method assumes that the input will meet the preconditions, so it does not include explicit error handling for invalid inputs.
    • Exception: None.
    • Special Value: None.

By understanding these aspects, you can grasp the overall design of the method and how it fits into solving the problem.

Problem Decomposition

  1. Problem Understanding:

    • The problem asks to find the maximum distance between valid index pairs (i, j) from two given non-increasing integer arrays nums1 and nums2. A pair is valid if i <= j and nums1[i] <= nums2[j].
  2. Initial Breakdown:

    • Major parts are:
      1. Input Validation.
      2. Finding valid pairs (i, j).
      3. Calculating the distance j - i for each valid pair.
      4. Keeping track of the maximum distance.
  3. Subproblem Refinement:

    • For Input Validation:
      1. Check if arrays are empty or null (though not strictly necessary based on the problem constraints).
    • For Finding valid pairs:
      1. Initialize pointers for nums1 and nums2.
      2. Iterate through both arrays to find valid pairs.
    • For Calculating the distance:
      1. Subtract i from j.
    • For Keeping track of maximum distance:
      1. Compare current distance with the maximum distance so far.
      2. Update maximum distance if needed.
  4. Task Identification:

    • Iterating through arrays and comparing elements are repeated tasks.
  5. Task Abstraction:

    • Each task is abstract enough to understand and does not require further breakdown for this problem.
  6. Method Naming:

    • validateInput: For validating the input arrays.
    • findValidPairs: For finding all valid pairs.
    • calculateDistance: For calculating the distance for a pair.
    • updateMaxDistance: For updating the maximum distance.
  7. Subproblem Interactions:

    • The order is: validateInput (optional), followed by findValidPairs, calculateDistance, and updateMaxDistance.
    • calculateDistance and updateMaxDistance are dependent on findValidPairs.
    • No real dependencies for other tasks; they can be part of a single main function or loop.

By understanding these elements, we get a clearer picture of the tasks at hand and how they interact to solve the problem.

From Brute Force to Optimal Solution

Brute Force Solution

The most straightforward approach is to consider every possible pair (i, j) where i is an index from nums1 and j is an index from nums2. For each pair, we check if it is a valid pair (i.e., i <= j and nums1[i] <= nums2[j]). If it is, we calculate the distance j - i and keep track of the maximum distance found so far.

Python Code for Brute Force

def max_distance_bruteforce(nums1, nums2):
    max_dist = 0
    for i in range(len(nums1)):
        for j in range(i, len(nums2)):
            if nums1[i] <= nums2[j]:
                max_dist = max(max_dist, j - i)
    return max_dist


  1. Time Complexity: The time complexity is (O(n \times m)), where (n) is the length of nums1 and (m) is the length of nums2. This is inefficient for large arrays.
  2. Space Complexity: The space complexity is (O(1)), which is fine, but the time taken is the issue here.

Optimized Solution

Step 1: Two Pointers

Both arrays are sorted in non-increasing order. We can take advantage of this by using two pointers. Start both pointers at the beginning of nums1 and nums2. Move the pointer in nums2 ahead until we find a valid pair or reach the end. Calculate the distance for each pair and keep track of the maximum.

Python Code after Step 1

def max_distance_optimized(nums1, nums2):
    max_dist = 0
    i, j = 0, 0
    while i < len(nums1) and j < len(nums2):
        if nums1[i] <= nums2[j]:
            max_dist = max(max_dist, j - i)
            j += 1
            i += 1
    return max_dist

Impact on Complexity

  1. Time Complexity: Reduced to (O(n + m)), which is much better than the brute-force approach.
  2. Space Complexity: Remains (O(1)), which is good.

By optimizing, we significantly reduced the time complexity while keeping the space complexity constant. We leveraged the sorted nature of the input arrays to accomplish this.

Code Explanation and Design Decisions

1. Initial Parameters

  • nums1 and nums2: These are the two input lists of integers. They represent the sequences of numbers where we want to find the maximum distance between indices i and j such that nums1[i] <= nums2[j] and i <= j.

2. Primary Loop

The primary loop iterates over both nums1 and nums2 using pointers i and j. Each iteration is an opportunity to find a pair (i, j) that satisfies the conditions of the problem.

3. Conditions Within the Loop

  • if nums1[i] <= nums2[j]: Checks if we have a valid pair. If it’s true, we compute the distance j - i.

The conditions are in place to ensure that we only consider pairs (i, j) that meet the problem’s constraints: nums1[i] <= nums2[j] and i <= j.

4. Updates Within the Loop

  • j += 1: We increase j to check the next index in nums2. The idea is to maximize the distance for a given i.
  • i += 1: We increase i because nums1[i] > nums2[j], and we cannot form a valid pair with this i and any further j.

These updates are necessary to progress through the lists and improve upon the maximum distance found so far.

5. Invariant

The invariant is that max_dist always contains the maximum valid distance found so far. This is ensured by the conditions within the loop and is crucial for the final answer.

6. Final Output

The final output is max_dist, the maximum distance between indices i in nums1 and j in nums2 that satisfy the problem’s conditions. It satisfies the problem’s requirements by giving the maximum valid distance.

By understanding these elements, we grasp not just what the code does, but why it works for solving this problem efficiently.

Coding Constructs

1. High-level Strategies

The code employs a Two Pointer Technique to traverse the two lists efficiently. It keeps track of the maximum distance between indices i and j where nums1[i] <= nums2[j] and i <= j.

2. Purpose to a Non-Programmer

This code finds the largest gap between two positions in two lists, such that the number at the first position is less than or equal to the number at the second position.

3. Logical Elements

  • Iteration: Looping through each element in the two lists.
  • Comparison: Checking if an element from the first list is less than or equal to an element from the second list.
  • State Maintenance: Keeping track of the largest gap found so far.

4. Algorithmic Approach in Plain English

  • Start at the beginning of each list.
  • Check if the number from the first list is smaller or equal to the one in the second list.
  • If it is, measure the gap between their positions and remember it if it’s the largest gap found so far.
  • Move to the next pair of numbers to repeat the process, advancing in a smart way to make the process faster.

5. Key Operations

  • The pointers i and j are moved based on comparisons between nums1[i] and nums2[j].
  • The distance between i and j is calculated and compared to the current maximum distance, updating it if a larger distance is found.

These steps are critical for solving the problem efficiently while adhering to the constraints and requirements.

6. Algorithmic Patterns

  • Two Pointer Technique: For efficient traversal of the lists.
  • Greedy Method: By always keeping the maximum distance found so far, the code ensures it ends up with the global maximum.

The algorithmic approach here is general and not tied to a specific programming language.

Language Agnostic Coding Drills

1. Distinct Coding Concepts

  1. Variable Initialization: Declaring variables to store temporary and final values.
  2. Array Traversal: Iterating over elements in an array or list.
  3. Element Comparison: Comparing values of elements in two different arrays.
  4. Two Pointer Technique: Using two pointers to traverse arrays in a coordinated manner.
  5. State Maintenance: Keeping track of the maximum value during traversal.
  6. Conditional Statements: Using if and else statements to guide logic.
  7. Loop Nesting: Placing one loop inside another, each having its unique role.
  8. Complexity Analysis: Understanding the time and space complexity of the algorithm.

2. Difficulty Classification and Description

  1. Variable Initialization
    Difficulty: Easy
    Reason: Basic step in any programming task.

  2. Array Traversal
    Difficulty: Easy
    Reason: Common operation in most coding problems.

  3. Element Comparison
    Difficulty: Easy-Moderate
    Reason: Introduces logic but is still straightforward.

  4. Conditional Statements
    Difficulty: Moderate
    Reason: Required for decision-making, a bit more complex.

  5. State Maintenance
    Difficulty: Moderate
    Reason: Must understand what needs to be stored and updated.

  6. Two Pointer Technique
    Difficulty: Moderate-Hard
    Reason: Requires coordinated movement of two pointers.

  7. Loop Nesting
    Difficulty: Hard
    Reason: Requires understanding the roles and coordination of multiple loops.

  8. Complexity Analysis
    Difficulty: Very Hard
    Reason: Needs a deeper understanding of algorithms and mathematics.

3. Problem-Solving Approach

  1. Understand the Problem: Break down the problem statement into specific requirements and constraints.

  2. Initialize Variables: Set up variables for storing temporary and final values.

  3. Simple Traversal: Begin by naively iterating over arrays to understand the nature of data.

  4. Element Comparison: Implement basic comparison logic between elements from two arrays.

  5. State Maintenance: Introduce a variable to hold the maximum value found so far.

  6. Conditional Logic: Incorporate if and else statements to decide when to update the maximum value and how to move the pointers.

  7. Two Pointer Technique: Replace naive traversal with two-pointer technique for efficiency.

  8. Loop Nesting: If the problem complexity requires, introduce nested loops for more elaborate traversal.

  9. Optimization: Continually check for areas of improvement in the existing logic.

  10. Complexity Analysis: Finally, evaluate the efficiency of your solution in terms of time and space complexity.

Each of these “drills” will build upon the previous one, contributing to the final, efficient solution.

Targeted Drills in Python

1. Python-based Coding Drills for General Concepts

Variable Initialization

# Initializing a variable to store an integer
num = 0

Array Traversal

# Traversing through an array
arr = [1, 2, 3, 4, 5]
for num in arr:

Element Comparison

# Comparing two numbers
a = 10
b = 20
if a > b:
    print("a is greater")
    print("b is greater")

Conditional Statements

# Using conditional statements
x = 5
if x > 0:
elif x == 0:

State Maintenance

# Maintaining a maximum value
max_val = 0
arr = [3, 4, 1, 7, 9]
for num in arr:
    if num > max_val:
        max_val = num
print("Maximum value is:", max_val)

Two Pointer Technique

# Two pointer technique
arr = [1, 2, 3, 4, 5]
start = 0
end = len(arr) - 1
while start < end:
    print(arr[start], arr[end])
    start += 1
    end -= 1

Loop Nesting

# Nested loops
for i in range(3):
    for j in range(3):
        print(i, j)

Complexity Analysis

This is more theoretical and doesn’t have a specific Python drill. It’s about understanding the time and space complexity of your code.

2. Problem-Specific Drills

None were specified since we don’t have a concrete problem statement. However, if there were domain-specific operations or complex data structures, this is where those drills would be created.

3. Integrating Drills into a Complete Solution

  1. Variable Initialization: Begin by initializing variables for storing temporary and final values.

  2. Array Traversal: Use a loop to iterate through given arrays.

  3. Element Comparison: Inside the loop, compare elements from different arrays.

  4. Conditional Statements: Use if and else statements to decide the logic based on comparison.

  5. State Maintenance: Introduce a variable to keep track of a state, like the maximum value.

  6. Two Pointer Technique: To optimize the traversal, implement the two-pointer technique.

  7. Loop Nesting: If the problem is complex, you might need to use nested loops.

  8. Complexity Analysis: Once all pieces are integrated, evaluate the time and space complexity of the code.

These individual drills can be plugged into a larger code structure in this order to construct the final solution. Each drill has its role and will contribute to solving the problem effectively.


Similar Problems

  1. Reverse Linked List: If you’re dealing with pointer manipulation or reversing data structures, this one can be quite similar.

  2. Palindrome Check: Whether for strings or numbers, this kind of problem often uses two-pointer techniques to compare elements from both ends toward the center.

  3. Maximum Subarray: This problem involves traversing an array to find contiguous elements with the maximum sum. It often involves state maintenance and element comparison.

  4. Merge Intervals: If your problem involves sorting and merging ranges, this problem is often quite similar in technique.

  5. Valid Parentheses: This problem usually involves stack data structures and element comparison, much like many other problems that involve checking for balanced symbols.

  6. Longest Common Prefix: This problem involves string or array comparison and could use a nested loop to iterate through each character of each string.

  7. Rotate Array: This problem often requires an understanding of array manipulation techniques, similar to those problems where you have to rearrange elements based on certain conditions.

  8. Single Number: Problems where you have to identify a unique element in a list often involve bitwise operations or other element comparison methods.

  9. Climbing Stairs: This is a simple dynamic programming problem. If your problem involves optimizing a straightforward recursive problem, this could be similar.

Each of these problems involves one or more of the underlying concepts like array traversal, element comparison, and state maintenance, making them related to the problem you are working on.