Minimum Time to Complete Trips
10 Prerequisite LeetCode Problems
This involves Binary Search and scheduling. Here are 10 problems to prepare:
704. Binary Search: A simple binary search problem to understand the basic concept.
278. First Bad Version: Another binary search problem but with a twist to understand how to apply the technique in different scenarios.
35. Search Insert Position: To understand how binary search can be used to find positions in an array.
852. Peak Index in a Mountain Array: This problem can help you understand how to modify the binary search algorithm to fit certain conditions.
540. Single Element in a Sorted Array: This problem is a good example of how binary search can be used in different scenarios.
378. Kth Smallest Element in a Sorted Matrix: This problem involves binary search and matrix manipulation which could be helpful.
658. Find K Closest Elements: An application of binary search in finding nearest elements.
668. Kth Smallest Number in Multiplication Table: This problem can help you understand how to perform binary search in a 2D space.
1095. Find in Mountain Array: This is an advanced binary search problem which will challenge your understanding of the concept.
875. Koko Eating Bananas: This problem is very similar to the one you want to solve. In both problems, you need to minimize the maximum amount of time.
Problem Analysis and Key Insights
What are the key insights from analyzing the problem statement?
Problem Boundary
What is the scope of this problem?
How to establish the boundary of this problem?
Problem Classification
The problem falls within the domain of scheduling and optimization. It involves analyzing the trips made by multiple buses and determining the minimum time required for them to complete a specified number of total trips.
What Components:
Input:
 An array of integers,
time
, wheretime[i]
represents the time taken by thei
th bus to complete one trip.  An integer,
totalTrips
, representing the total number of trips all buses should make.
 An array of integers,
Output:
 An integer representing the minimum time required for all buses to complete at least
totalTrips
trips.
 An integer representing the minimum time required for all buses to complete at least
Constraints:
 Limits on the length of
time
and the values oftime[i]
andtotalTrips
.
 Limits on the length of
Objective:
 Determine the minimum time required for all buses to complete the specified number of total trips.
This can be classified as a scheduling problem, where the task is to coordinate the trips of several buses to minimize the total time required to complete a given number of trips. The independent and successive operation of buses introduces additional complexity into the problem.
The categorization as a scheduling problem suggests that techniques and algorithms used in operations research, optimization, and time management may be applicable to solve the problem. Understanding the frequency and timing of each bus’s trips, and how these interact to contribute to the total number of trips, will be key to developing a solution.
Distilling the Problem to Its Core Elements
Fundamental Concept or Principle: The fundamental concept this problem is based upon is the principle of scheduling and optimization. It’s about efficiently allocating time for buses to complete a given number of trips in the minimum amount of time.
Simplest Way to Describe the Problem: Imagine a fleet of buses where each bus takes a specific amount of time to complete a trip and can start the next trip immediately. You need to figure out the least amount of time it takes for all the buses combined to complete a certain number of trips.
Core Problem: The core problem is to find the minimum time required for a fleet of buses to collectively complete a given number of trips.
Simplifying the Problem Statement: How quickly can a set of buses, each with a known travel time for one trip, complete a specified total number of trips?
Breaking Down the Problem into Key Components:
 Understanding Each Bus’s Schedule: Determine how many trips each bus can complete within a given time frame.
 Total Trips Calculation: Sum the trips completed by all buses at different time intervals to track the cumulative trips.
 Finding the Minimum Time: Identify the minimum time at which the total number of trips is greater than or equal to the target.
Minimal Set of Operations:
 Initialize Variables: Set up variables to track the current time and total trips.
 Iterate through Time: Increment the time step by step, or by using a more efficient search like binary search.
 Calculate Trips at Each Time: For each time step, calculate the number of trips each bus has completed.
 Check Total Trips: Sum the trips from all buses and compare to the target total trips.
 Identify Minimum Time: Once the total trips meet or exceed the target, record the current time as the minimum required time.
By analyzing the buses’ schedules and how they contribute to the total trips over time, and by methodically checking the total trips against the target, these operations provide a path to solving the problem.
Visual Model of the Problem
Visualizing the problem statement can greatly aid in understanding and solving the problem. Here’s how you might visualize this problem:
Timeline Visualization:
 Draw a horizontal timeline representing time in units (e.g., hours or minutes).
 Above the timeline, sketch lines or bars for each bus, showing how many trips it completes at various time intervals.
 Below the timeline, you can plot a graph representing the cumulative number of trips completed by all buses over time.
 Mark the point on the timeline where the cumulative trips reach or exceed the target number, which will represent the minimum time required.
Table Visualization:
 Create a table where each row represents a bus, and each column represents a unit of time.
 Fill in the cells with the number of trips each bus has completed at that time.
 Sum the trips at each time step in a row at the bottom of the table to see how the total trips build up over time.
Graph Visualization:
 Create a graph with time on the xaxis and the total number of trips on the yaxis.
 Plot points for each time unit, showing the cumulative number of trips at that time.
 Draw a line or curve through the points to see how the total trips increase over time.
 Mark the point where the line meets or exceeds the target number of trips, highlighting the minimum time required.
By choosing one of these visualizations, or even a combination of them, you can create a visual representation of how the buses’ trips accumulate over time, and how that relates to the total trips required. This can make it easier to conceptualize the problem and identify the minimum time needed for all buses to complete the desired number of trips.
Problem Restatement
You have a list of buses, each with a specific time it takes to complete a trip. Every bus can make as many consecutive trips as needed, and they all operate independently of each other. Given the time taken by each bus for a single trip and a target number of total trips that all buses together should complete, you need to find the minimum time required to reach or exceed this target number of trips.
In simpler terms:
 You have several buses.
 Each bus takes a certain amount of time to complete one trip.
 All buses can make multiple trips, one after the other.
 You need to find out the least amount of time required for all buses combined to complete a specified total number of trips.
The constraints include:
 The number of buses is between 1 and 105.
 The time taken by each bus and the total target trips are between 1 and 107.
The problem focuses on optimizing time by understanding how the cumulative trips build up, given the individual trip times for each bus.
Abstract Representation of the Problem
An abstract representation of a problem helps to emphasize the core structure and relationships without getting lost in specific details. Here’s how we might describe this problem in an abstract way:
 Elements: Consider a set of entities, each having an associated cost or time value. These entities can repeatedly perform a unit task, and each repetition takes the same amount of time.
 Goal: Determine the minimum total time required for the entities to collectively perform a given number of unit tasks.
 Constraints:
 The time required by each entity to perform the unit task is fixed.
 The entities can perform the unit task concurrently and can repeat the task immediately without delay.
 The total number of tasks to be performed is a given target.
By stripping away the realworld context (buses, trips, etc.), we can see that the problem essentially revolves around optimizing a cumulative sum of repetitive actions carried out by multiple entities, each having a specific rate of action. This abstract representation could apply to various realworld scenarios beyond the one described, emphasizing the universal structure of the problem.
Terminology
While the problem itself is stated in relatively simple terms, there are some underlying concepts that might be considered technical:
Time Complexity: It refers to the amount of time an algorithm takes to complete as a function of the length of the input. In this problem, considering the time complexity of the solution is crucial to ensure that the algorithm is efficient, especially given the constraints.
Concurrency: This concept is implicit in the problem, as all buses operate independently and concurrently. Concurrency refers to the execution of multiple processes at the same time. Here, it implies that all buses can make trips simultaneously.
Optimization Problem: The problem aims to find the best (minimum) time for a given condition. Optimization problems seek the best solution according to specific criteria and constraints. In this context, it’s about minimizing the time needed for all buses to complete a certain number of trips.
Repetitive Task: A task that can be performed multiple times in succession. In this problem, the task is the completion of a trip by a bus, and it can be repeated immediately after finishing the previous one.
Divide and Conquer: This is a potential strategy for solving the problem where you divide the problem into smaller subproblems, solve those, and combine their solutions to solve the original problem. This could be applied to finding the minimum time by dividing the total trips among buses and conquering them individually.
Understanding these concepts and terms helps in grasping the nature of the problem and the approaches that might be applicable to solving it.
Problem Simplification and Explanation
Let’s break down the problem into simpler terms and use an analogy to understand it better.
Problem Breakdown
 Buses: We have a fleet of buses, each taking a different amount of time to complete one trip.
 Trips: Each bus can make multiple trips, one after the other, without any break.
 Total Trips: We have a target number of total trips that need to be completed by all buses combined.
 Goal: We want to find the minimum time required for all buses to complete the target number of total trips.
Metaphor/Analogy
Imagine you are managing a team of painters, and each painter paints at a different speed. You have a certain number of walls that need to be painted, and you want to find out how long it will take to finish painting all the walls if the painters work simultaneously.
 Painters: Correspond to the buses in the problem.
 Walls: Correspond to the total trips that need to be completed.
 Painting Speed: Correspond to the time taken by each bus to complete one trip.
 Total Painting Time: Correspond to the minimum time required for all buses to complete the target number of total trips.
Interaction of Concepts
 Concurrency: All buses (painters) operate simultaneously, working independently of one another.
 Repetition: Each bus (painter) can complete a trip (paint a wall) and then immediately start another, continuing this pattern.
 Optimization: We are looking to find the shortest amount of time in which all required trips (walls) can be completed.
By thinking of the buses as painters, each working at their own pace to complete a common goal, you can visualize the problem more concretely and understand how the key concepts interact with one another.
Constraints
Identifying specific characteristics or conditions can help us in crafting an efficient solution for the problem. Here’s what can be exploited:
Sorted Time Array: The
time
array, which denotes the time taken by each bus to complete one trip, can be sorted. By sorting this array, we may have a better understanding of how buses are completing trips over time and can efficiently calculate how many trips are being completed at any given time point.Repetitive Trips: Since each bus can make multiple trips successively, and the next trip starts immediately after completing the current one, the time taken by a bus to complete trips will repeat in cycles. This can help us optimize calculations over a period.
Simultaneous Operation: All buses are operating simultaneously and independently. This means we can analyze each bus’s performance separately and then combine the results, possibly leading to parallel or divideandconquer strategies.
Large Range of Values: With
time[i]
andtotalTrips
ranging up to (10^7), brute force or purely iterative approaches may not be feasible. Recognizing this helps us avoid those paths and look for a more efficient strategy like binary search or mathematical manipulation.Uniqueness of Trips: Each trip by a bus is distinct and contributes to the total trip count. There is no interdependence between buses, so we can isolate each bus’s effect on the total trip count. This might lead us to use mathematical formulas to derive the contribution of each bus to the total trips at a given time.
These characteristics and constraints can guide our approach, helping us focus on efficient algorithms that take advantage of the specific patterns and numerical ranges provided in the problem.
Analyzing the constraints of the problem leads us to several key insights:
Simultaneous Operation: All buses operate simultaneously and independently. This insight tells us that we can break down the problem into parts, analyzing each bus’s contribution to the total trips separately.
Repetitive Nature: Since each bus’s trips repeat after completing one cycle (equal to the time taken for one trip), there’s a periodic pattern in their behavior. Recognizing this pattern can lead us to avoid unnecessary repetitive calculations.
Large Range of Values: The constraints on the values of
time[i]
andtotalTrips
(up to (10^7)) indicate that a naive or bruteforce approach is likely to be infeasible. This encourages us to seek more efficient algorithms like binary search or mathematical formulas that handle large values efficiently.Possibility of Sorting: If needed, sorting the
time
array could be an option, as it would still be within the computational constraints. This could potentially lead to a more efficient calculation of the trips made by all buses.Uniqueness of Trips: The buses’ independent operation and distinct trips help us understand that we can treat each bus separately and then combine the results. This may lead to solutions that break down the problem into smaller, more manageable parts.
Potential for Mathematical Approach: The repetitive and independent nature of the buses’ trips hints at the possibility of a mathematical approach. Understanding the cycles and behavior of each bus might lead to an equation or formula to represent the trips, thereby simplifying calculations.
These insights derived from the constraints guide our approach towards the problem and help us avoid paths that are likely to be less efficient. By recognizing the specific characteristics of the problem, we can focus on strategies that exploit these features to arrive at an optimal solution.
Case Analysis
Here are additional examples or test cases, categorized to cover different aspects of the problem:
1. Single Bus Case
Input: time = [4], totalTrips = 2
Output: 8
Explanation: Since there’s only one bus, the total time required to complete 2 trips is 8. It represents the simplest scenario, with the minimum constraint values.
2. Multiple Buses with Same Time
Input: time = [3,3,3], totalTrips = 9
Output: 3
Explanation: All buses have the same time and can complete 9 trips in 3 hours. This case emphasizes that if the buses have identical time intervals, the total trips can be calculated simply.
3. Multiple Buses with Different Time
Input: time = [1,2,3], totalTrips = 5
Output: 3
Explanation: This case showcases the complexity when different buses have varying time intervals for a trip. Understanding how these intervals interact is key to solving this.
4. Single Trip Required
Input: time = [2,4,6], totalTrips = 1
Output: 2
Explanation: This edge case shows the scenario where only one trip is required. The minimum time from the array is the output. This special case could be a pitfall if not handled correctly.
5. All Buses Have Maximum Time
Input: time = [10^7, 10^7, 10^7], totalTrips = 10^7
Output: 10^7
Explanation: This scenario checks the boundaries of the given constraints and ensures that the solution can handle large numbers.
6. TotalTrips is a Multiple of Time Sum
Input: time = [2, 3], totalTrips = 10
Output: 4
Explanation: This example highlights the scenario where the totalTrips is a multiple of the sum of the time array. It can help in understanding the cyclic nature of the problem.
Edge Cases Summary
 A single bus represents the minimum constraint.
 Scenarios where totalTrips is 1 or a multiple of the sum of the time array.
 Scenarios with maximum constraints, e.g., all buses have maximum time and the maximum totalTrips.
These cases collectively illustrate the range of possible scenarios, including both typical and extreme situations. Analyzing these helps in understanding the different dimensions of the problem and ensures that the solution is wellrounded and robust.
Analyzing the different cases provides several key insights into the problem:
Uniform vs. Varied Bus Time: If all buses take the same amount of time for a trip, the problem simplifies, as the total time will be a straightforward calculation. However, if the buses have varied times, understanding their interactions and cumulative effect becomes crucial.
Minimum Time: The minimum time among all the buses will always be a lower bound for the solution. If only one trip is needed, the minimum time is the solution.
Cyclic Nature: The problem exhibits a cyclic nature, where the pattern of trips completed repeats after a sum of the time array. If totalTrips is a multiple of this sum, understanding this cycle can lead to a more efficient solution.
Handling Extremes: Edge cases like having a single bus, all buses taking the maximum time, and the largest totalTrips help in understanding how the solution should behave at the boundaries of the constraints.
Incremental Approach: The way buses complete trips incrementally can be exploited to calculate the total time in an optimized manner. Instead of simulating each step, understanding how the times interact can lead to a quicker calculation.
Division and Modulo Operations: Insights from the cyclic nature and the incremental approach might lead to the idea of using division and modulo operations to find the solution more efficiently.
These insights collectively contribute to a deeper understanding of the problem’s nature and structure, guiding the design of a robust and efficient solution that caters to all possible scenarios.
Identification of Applicable Theoretical Concepts
Several mathematical and algorithmic concepts can be applied to simplify the problem and make it more manageable:
Binary Search: Since we’re looking for the minimum time required for all buses to complete a given number of trips, we could employ a binary search on the time to efficiently find the answer. This reduces the complexity from a bruteforce approach.
Mathematical Modelling: Understanding the cyclic pattern formed by the buses completing their trips can be modeled mathematically. Utilizing division and modulo operations, we can calculate how many full cycles are needed, and what remains to be covered. This provides an efficient way to calculate the total time.
Greatest Common Divisor (GCD) and Least Common Multiple (LCM): The interaction between different time intervals might be explored using GCD and LCM, as they provide a mathematical way to understand the relationship between different numbers.
Simulation with Optimization: Instead of simulating each step, we can increment the time by the smallest time interval in the array and calculate the total trips completed at each step. This uses the idea of an incremental approach to reach the solution more efficiently.
Greedy Approach: By prioritizing the buses that take the least amount of time, we might design a greedy algorithm that builds the solution iteratively, ensuring at each step that the choice made is the best one to reach the optimal solution.
Complexity Analysis: Understanding the constraints and complexity of the problem helps to pick the right algorithm. In this case, the constraints suggest that a more efficient algorithm like binary search would be a viable solution.
Handling Edge Cases: Mathematical reasoning to handle edge cases, such as a single bus or all buses taking the same amount of time, can provide shortcuts to the solution in these special scenarios.
Applying these concepts and methodologies can turn a seemingly complex problem into a more manageable and logically tractable task, guiding the way to an efficient solution.
Simple Explanation
Let’s think of this problem in terms of a fleet of buses running on different routes. We’ll use a simple metaphor and everyday language to make it accessible to someone without a technical background.
Imagine you have several buses, and each bus takes a different amount of time to complete one trip around its route. Some buses are faster and can finish their route in just a few minutes, while others are slower and take longer.
Now, you’re given a total number of trips that all the buses together must complete. You want to figure out the shortest amount of time it will take for the buses to make these trips.
Here’s an everyday example to illustrate the idea:
 Think of three buses: one takes 1 minute for a trip, the second takes 2 minutes, and the third takes 3 minutes.
 You need to find the minimum time for all buses to complete at least 5 trips in total.
The problem is like coordinating a team of runners with different speeds in a relay race. You need to figure out when the whole team will finish a certain number of laps, considering each runner’s pace.
In layman’s terms, you’re trying to find the quickest way to complete a task by coordinating multiple participants, each with a different speed, to reach a common goal. It’s like a puzzle where you fit the pieces together to find the most efficient way to achieve the objective.
Problem Breakdown and Solution Methodology
Let’s break down the process into smaller, more manageable steps to solve this problem and use the metaphor of buses completing trips to guide our understanding.
StepbyStep Approach
Understand the Constraints: First, we recognize that each bus has its unique trip time, and we need to figure out the minimum time for all buses to complete a given total number of trips.
Calculate Individual Contributions: Determine how many trips each bus can complete in a given time. For example, if the bus takes 3 minutes per trip, in 6 minutes, it would have completed 2 trips.
Binary Search for Minimum Time: We can perform a binary search to find the minimum time required for all buses to complete the total number of trips. This is like testing different times and checking if the total number of trips completed is enough.
a. Start and End Points: Choose a start point as the minimum trip time and an end point as the maximum possible time (e.g., maximum trip time multiplied by totalTrips).
b. Midpoint Calculation: Calculate the midpoint between the start and end points and determine the total number of trips completed by all buses in that time.
c. Check Condition: If the total number of trips is less than totalTrips, update the start point to be the midpoint + 1. If it is greater or equal, update the end point to be the midpoint.
d. Repeat: Repeat steps bc until the start point is greater than the end point. The end point will be the minimum time required.
Final Result: The result from step 3 will give the minimum time required for all buses to complete the total number of trips.
Effects of Changes in Parameters
 Increasing Total Trips: If the total number of trips required increases, the minimum time will likely increase, as more trips need to be completed.
 Changes in Trip Times: If the trip time for a specific bus changes, it may affect the overall minimum time, depending on whether the change makes the bus faster or slower.
Example Case
 time = [1,2,3], totalTrips = 5
 Step 1: Understand the problem.
 Step 2: Recognize that each bus completes a different number of trips in a given time.
 Step 3: Perform binary search.
 Start: 1, End: 15
 Mid: 8, Total Trips: 8 (1x8 + 2x4 + 3x2) > End becomes 8
 Mid: 4, Total Trips: 4 (1x4 + 2x2 + 3x1) > Start becomes 5
 Mid: 6, Total Trips: 6 (1x6 + 2x3 + 3x2) > Start becomes 7
 Mid: 7, Total Trips: 7 (1x7 + 2x3 + 3x2) > Start becomes 8
 Step 4: The final result is 8, but since the problem requires the last time when the buses complete the totalTrips, the answer is 3.
The approach leverages the sorted nature of the trip times and exploits the characteristics of the problem to find an efficient solution. By breaking down the problem into smaller parts and applying a binary search, we create an intuitive and effective method for solving the problem.
Inference of ProblemSolving Approach from the Problem Statement
Here are the key terms or concepts in the problem and how they guide the approach to solving it:
Array of Time (time[i]): This represents the time taken by the ith bus to complete one trip. It’s the fundamental data we work with, guiding us to calculate individual contributions and understand how different buses complete trips at different rates.
TotalTrips: The total number of trips all buses must complete. This sets the target we are trying to reach and defines the goal of finding the minimum time to meet or exceed this target.
Multiple Successive Trips: Each bus can make multiple trips successively without delay. This informs the approach by letting us understand that the number of trips a bus can make is directly proportional to the time given, guiding the use of a binary search method to find the minimum time.
Independent Operation of Buses: Each bus operates independently, meaning we can calculate the trips for each bus separately and sum them up. This independence allows us to simplify the problem and consider each bus’s contribution individually.
Minimum Time Required: The goal is to find the minimum time needed for all buses to complete at least a total number of trips. This objective steers us towards an optimization strategy, and in this context, binary search is a suitable method to efficiently find the minimum time.
Constraints: The constraints of 1 <= time[i], totalTrips <= 10^7 emphasize that the solution must be efficient. This informs the choice of an efficient algorithm like binary search, which operates in logarithmic time.
Binary Search: The sorted nature of time, the need to find a minimum, and the constraints lead us to the binary search strategy. By testing midpoints and adjusting ranges, we can efficiently locate the minimum time required.
By identifying and understanding these key terms and concepts, we create a clear roadmap for solving the problem, choosing strategies that align with the problem’s structure, requirements, and constraints.
The use of binary search in the given problem can be inferred through several observations and key insights:
Monotonic Function: The relationship between time and total trips is a monotonic function. As time increases, the number of trips that the buses can complete also increases. This property ensures that there is a specific time value for which the number of trips equals or exceeds the required total trips, and we can use binary search to find this value.
Search Space: We have a clear search space for the minimum time required, ranging from the minimum time taken by any bus to the maximum time it would take for all buses to complete the total trips. This search space is continuous and ordered, making it suitable for binary search.
Goal of Finding a Specific Value: The problem is to find the minimum time required for the buses to complete a specific number of trips. This is a search problem where we’re looking for a specific value that meets certain conditions, and binary search is an efficient algorithm for such problems.
Efficiency Concerns: A brute force or linear search would be inefficient in this case, especially if the search space is large. Recognizing that a more efficient solution is needed, and that the problem exhibits the characteristics suitable for binary search, leads to the choice of this method.
In summary, the properties of the problem, such as the monotonic relationship between time and total trips, the clear and ordered search space, and the goal of finding a specific value, all contribute to the insight that binary search can be applied to solve this problem efficiently.
Simple Explanation of the Proof
Let’s break down the algorithm and its proof into simple terms to understand how and why it works. We’ll use the example of finding the minimum time required for buses to complete a total number of trips.
Understanding the Problem: We have a list of times that different buses take to complete a trip, and we want to find the minimum time needed for the buses to complete a certain total number of trips.
Binary Search Basics: Binary search is a method used to find a specific value within an ordered search space. It repeatedly divides the search space in half until it finds the desired value.
Applying Binary Search:
 We start with the minimum and maximum possible time values. The minimum value could be the smallest time taken by any bus, and the maximum could be the time taken to complete all the total trips.
 We find the middle value and check if the total trips completed in that time are greater or equal to the required total trips.
 If it’s greater or equal, we know that the minimum time lies in the left half of the search space, so we update our maximum time to the middle value.
 If it’s less, the minimum time lies in the right half, so we update our minimum time to the middle value plus one.
 We continue this process until the search space is narrowed down to one value, which is our answer.
Proof Explanation:
 Correctness: We are consistently reducing the search space while ensuring that the desired value is within that space. The monotonic property ensures that once we find a time value that meets the requirement, all greater values will also meet it, so we can safely narrow our search.
 Termination: The search space keeps getting halved, so we are guaranteed to reach a point where it consists of only one value, and the algorithm terminates.
 Efficiency: By dividing the search space in half at each step, we ensure that the algorithm runs in logarithmic time, making it highly efficient.
Easy Analogy: Think of it as looking for a specific word in a dictionary. Instead of going through every page, you open the dictionary in the middle and check if the word is before or after the page you’re on. You then repeat this process with the corresponding half of the dictionary until you find the word.
By following this process, the binary search algorithm ensures that it finds the desired value efficiently and correctly. The logic and structure of the algorithm guarantee that it will converge on the correct answer in logarithmic time.
Stepwise Refinement
Let’s break down the problem into more detailed and actionable steps, identify independent parts, and recognize repeatable patterns:
Stepwise Refinement of the Approach
Initialize Boundaries for Binary Search: Set low as 0 and high as the maximum possible time (e.g., the maximum time in the array times the totalTrips).
Implement Binary Search: a. While low is less than or equal to high: b. Calculate mid as the midpoint between low and high. c. Determine the total trips made by all buses in ‘mid’ time. d. If the total trips are less than totalTrips, set low to mid + 1. e. Otherwise, set high to mid  1.
Return the Result: The final value of low is the minimum time required for all buses to complete the totalTrips.
Granular, Actionable Steps
 Calculating Total Trips in ‘mid’ Time:
a. Initialize a variable total_trips to 0.
b. Loop through each bus’s time:
 Divide ‘mid’ by the bus’s time to find how many trips that bus can make.
 Add the result to total_trips. c. Compare total_trips with totalTrips to decide the next step in the binary search.
Independent Parts of the Problem
Calculating Total Trips for Given Time: The part where we calculate how many trips can be made by each bus in a given time is independent. It can be extracted into a separate function and tested separately.
Binary Search Algorithm: The core binary search logic can be implemented and tested independently from the specific context of this problem.
Repeatable Patterns
Binary Search Pattern: The binary search pattern is a classic repeatable pattern used in this problem to efficiently find the minimum time.
Calculating Trips for a Bus: The logic to calculate how many trips a bus can make in a given time is repeated for each bus. It’s a repeatable pattern within the loop.
By refining the approach into these detailed steps, independent parts, and recognizing repeatable patterns, we create a clear and structured path to implement the solution, allowing for potential code reusability and easier debugging.
Solution Approach and Analysis
Let’s break down the solution and use metaphors to explain each step:
Problem
You are given an array of times taken by buses for one trip and a total number of trips required. You need to find the minimum time required for all buses to complete the given total number of trips.
Approach
Imagine the buses as factory machines, each producing a widget in a specific time. You need to find the minimum time needed to produce a required number of widgets.
Steps
Set Initial Boundaries for Search:
 Start with the minimum time as 0 (no time passed) and the maximum time as the time taken by the slowest bus multiplied by the total number of trips required.
Implement a Binary Search to Narrow Down the Time:
 Metaphor: It’s like cutting a cake into two equal parts and then examining which part has the piece you are looking for.
Calculate Total Trips for a Given Time (mid):
 Loop through each bus’s time:
 Determine how many trips that specific bus can make in ‘mid’ time.
 Sum these trips for all buses.
 Metaphor: Imagine each bus as a worker in a factory, and you are calculating how many widgets they can make in ‘mid’ hours.
 Loop through each bus’s time:
Adjust the Search Boundaries Based on the Total Trips:
 If total trips in ‘mid’ time are less than required, increase the minimum time.
 If total trips are equal to or more than required, decrease the maximum time.
 Repeat steps 3 and 4 until the minimum time exceeds the maximum time.
Return the Final Minimum Time: The value where the search ended is the minimum time required.
Effects of Changes in Parameters
 Increasing TotalTrips: Will likely increase the minimum time needed.
 Adding Faster Buses: Can decrease the minimum time as more trips can be made in the same time.
 Increasing Time of Individual Bus: May increase the minimum time, especially if it’s the fastest bus.
Example Case
 Input: time = [1,2,3], totalTrips = 5
 Output: 3
Working
 Initialization: low = 0, high = 15
 First Iteration: mid = 7.5, total trips = 7 + 3 + 2 = 12 (more than 5), so high = 7
 Second Iteration: mid = 3.5, total trips = 3 + 1 + 1 = 5 (equal to 5), so high = 2
 Third Iteration: low now exceeds high, so the answer is low = 3
The approach leverages the sorted and independent nature of the buses’ times, using binary search to efficiently find the minimum time required for the given total number of trips.
Identify Invariant
In computer science, an invariant is a condition that can be relied upon to be true during the execution of a program, or during some portion of it.
In the context of this problem, the invariant we maintain is that the solution (i.e., the minimum time required for all buses to complete the given total number of trips) always lies within the range defined by our current low and high bounds during the binary search.
At every step of our binary search, we adjust either the low or high bound based on the total trips made in the mid time. Our invariant is that the correct minimum time must be within these bounds, and we never move the bounds outside the range where the correct answer could be.
This invariant guides our algorithm and ensures its correctness as it narrows down the possible solution space to converge on the correct answer. It also ensures that we do not overlook any possible solutions within the defined range during our search.
Identify Loop Invariant
A loop invariant is a property or condition that holds true before and after each iteration of a loop. Understanding the loop invariant can be critical to understanding the correctness of an algorithm.
In the context of this problem, if we are using a binary search algorithm to find the minimum time required for all buses to complete the given total number of trips, the loop invariant would be something like:
At the beginning of every iteration of the binary search loop, the correct minimum time required lies within the current range defined by the low and high bounds.
The value of the low bound is never greater than the actual minimum time required, and the value of the high bound is never less than the actual minimum time required.
This loop invariant ensures that the algorithm will eventually converge to the correct answer because, with each iteration, we narrow the range without excluding the actual solution. By maintaining this invariant, we ensure the correctness of the algorithm and guarantee that it will find the desired minimum time required.
In this problem, the concept of a loop invariant and an invariant may refer to different things.
Loop Invariant: A loop invariant is a property or condition that holds before and after each iteration of a loop. In the binary search portion of the code above, a loop invariant might be that the correct answer exists within the current
[left, right]
interval. This property must be true before the loop begins, remain true after each iteration, and still be true when the loop terminates.Invariant: An invariant, more generally, refers to a property or condition that remains unchanged throughout the execution of a program or within a particular scope. It doesn’t necessarily have to relate to a loop. For example, the fact that the
time
array andtotalTrips
variable remain constant throughout the code’s execution could be considered invariants of this problem.
So, in this problem, the terms are different, referring to distinct concepts related to the problem’s constraints and the algorithm’s properties.
Thought Process
The problem involves an array of bus trip times and a total number of trips that need to be completed. The goal is to find the minimum time required for all buses to complete at least the specified number of total trips. Here’s the thought process:
Identify the Range: The maximum time required can be calculated as the time of the slowest bus multiplied by the total number of trips. The minimum time can be considered as 0.
Binary Search Approach: We can use binary search within this range to find the minimum time required. By checking the midpoint of the range and calculating the total trips that can be completed in that time, we can narrow down the search space.
Calculate Total Trips: In the binary search, for each midpoint time, we calculate the total trips completed by all buses up to that point. If the total is less than the required trips, we need more time, so we search in the higher half. If the total is greater or equal, we can potentially do it in less time, so we search in the lower half.
Finalize the Result: Continue the binary search until the search space is minimized to a point, and we find the minimum time required to complete the specified total trips.
Python Code


Insights
 The problem suggests a binary search approach due to the sorted nature of the bus trip times and the search for a minimum time within a range.
 By considering the time required for the slowest bus and the total trips, we can efficiently narrow down the search space and find the optimal solution.
 This binary search is applied on time, not on the bus trip times array, which makes the problem unique and interesting.
Example Usage


Establishing Preconditions and Postconditions
Parameters:
 Inputs:
time
(List of integers) andtotalTrips
(integer).  Types:
time
is a List of integers, andtotalTrips
is an integer.  Representation:
time
represents the time taken by each bus to complete one trip, andtotalTrips
represents the total number of trips that all buses must complete.
 Inputs:
Preconditions:
 The
time
array must have at least one element, and all values must be greater than or equal to 1.  The
totalTrips
must be greater than or equal to 1.  The program expects the
time
array andtotalTrips
to meet the specified constraints.
 The
Method Functionality:
 The method is expected to calculate the minimum time required for all buses to complete at least
totalTrips
trips.  It uses the
time
array andtotalTrips
to calculate this minimum time.
 The method is expected to calculate the minimum time required for all buses to complete at least
Postconditions:
 The state of the program remains unchanged; no side effects on input parameters.
 The return value represents the minimum time required for all buses to complete the specified number of trips.
 No side effects are expected from this method.
Error Handling:
 The method assumes that the preconditions are met and does not explicitly include error handling for scenarios where the input does not meet the constraints.
 If the input parameters do not meet the specified constraints, the behavior of the method might be undefined. It could return an incorrect result or encounter a runtime error, depending on the actual implementation. There is no indication in the problem statement that exceptions should be thrown or special values returned for invalid inputs.
Problem Decomposition
Problem Understanding:
 The problem involves an array of buses, with each element representing the time a bus takes to complete a trip. All buses can make trips concurrently, and the goal is to find the minimum time required for all buses to complete a given total number of trips.
Initial Breakdown:
 Determine how many trips each bus can make in a given time.
 Calculate the total trips made by all buses at different time intervals.
 Find the minimum time required to reach or exceed the total number of desired trips.
Subproblem Refinement:
 For determining the number of trips each bus can make:
 Divide the given time by the time each bus takes to complete a trip.
 For calculating total trips:
 Sum the trips made by each bus.
 For finding minimum time:
 Utilize a binary search or iterative approach to find the least time needed.
 For determining the number of trips each bus can make:
Task Identification:
 Calculating the number of trips made by a bus in a given time.
 Summing the total trips made by all buses.
 Determining the minimum time required.
Task Abstraction:
 Each task is abstracted to work on generic input, but within the context of the problem.
Method Naming:
calculateTripsForBus
: Determines the number of trips made by a bus in a given time.calculateTotalTrips
: Sums the total trips made by all buses.findMinimumTime
: Determines the minimum time required to reach the total number of desired trips.
Subproblem Interactions:
 First, determine how many trips each bus can make in a given time.
 Then, calculate the total trips made by all buses at different time intervals.
 Finally, find the minimum time that results in the total number of desired trips being met or exceeded.
 The tasks interact with each other, where the calculation of trips for each bus feeds into the calculation of total trips, which then informs the search for the minimum time. The tasks need to be performed in this order, as each builds on the results of the previous.
From Brute Force to Optimal Solution
Let’s begin by understanding the brute force solution and then move on to optimizing it.
Brute Force Solution
 Iterate over a time variable starting from the minimum time a bus takes to complete a trip, and increment it by the same minimum time on each iteration.
 For each time value, calculate how many trips each bus can make in that time and sum them up to get the total trips made by all buses.
 If the total trips meet or exceed the required total trips, return the current time value as the minimum time required.
Code Snippet (Brute Force)


Inefficiencies in the Brute Force Solution
 This solution is highly inefficient as it iterates over the time in small increments, and it can take a large number of iterations to reach the desired total trips.
 Time Complexity: O(T * N), where T is the total trips and N is the number of buses.
 Space Complexity: O(1)
Optimization
 Binary Search Approach: Instead of incrementing the time linearly, we can perform a binary search over the time to find the minimum time required more efficiently.
Optimized Code


How Optimization Impacts Time and Space Complexity
 By using binary search, we reduce the time complexity to O(N * log(T)), where N is the number of buses and T is the maximum time required for all buses to complete the total trips.
 The space complexity remains the same, O(1), as we are still using a constant amount of extra space.
The optimized solution leverages a binary search approach to efficiently find the minimum time required for all buses to complete the desired total trips. It greatly improves the time complexity over the brute force solution, while maintaining the same space complexity.
Code Explanation and Design Decisions
We’ll continue with our example of finding the minimum time required for buses to complete a total number of trips, using binary search, and explain the rationale behind each part of the solution.
Initial Parameters:
time
: List of integers representing the time taken by different buses for a single trip.totalTrips
: Integer representing the total number of trips that must be completed. These parameters define our problem’s constraints and guide the solution.
Primary Loop/Iteration:
 The binary search loop iterates over the possible time values to find the minimum time required.
 Each iteration narrows down the search space, thus getting closer to the solution.
Conditions/Branches Within the Loop:
 We calculate the middle value between the minimum and maximum time and check if the total trips completed in that time are greater or equal to
totalTrips
.  If true, it means the solution could lie in the left half, and we update the maximum time to the middle value.
 If false, the solution is in the right half, and we update the minimum time. This branching is based on the monotonic property of the problem, where once a time value meets the requirement, all greater values will too.
 We calculate the middle value between the minimum and maximum time and check if the total trips completed in that time are greater or equal to
Updates/Modifications to Parameters:
 The modifications to the minimum and maximum time values during each iteration are essential for narrowing down the search space.
 They reflect changes in our understanding of where the solution lies, based on the logic of the problem.
Invariant:
 The invariant is that the correct minimum time always lies within the current search space (between the current minimum and maximum time).
 This helps ensure that we don’t lose track of the solution and guarantees the correctness of the algorithm.
Significance of the Final Output:
 The final output represents the minimum time required for the buses to complete the given total number of trips.
 It satisfies the problem’s requirements by finding the optimal solution according to the given constraints.
The solution leverages the binary search algorithm to efficiently find the minimum time required. The iterations, branching, modifications, and invariant all work together within the context of the problem’s constraints to guide the algorithm to the correct and optimal solution.
Coding Constructs
HighLevel Strategies:
 The code uses a binary search algorithm, which is a divideandconquer strategy to efficiently search through sorted data by repeatedly dividing the search space in half.
Explanation to a NonProgrammer:
 Imagine you’re trying to find a specific page in a sorted book. Instead of going page by page, you open the book in the middle and check whether your page comes before or after that point. Depending on your check, you then only look at the half of the book where the page must be. Repeat this process, and you’ll find your page quickly.
Logical Elements:
 Loop (to iterate through the search space)
 Conditional Statements (to decide the direction of search)
 Calculations (to compute the middle value and total trips)
Algorithmic Approach in Plain English:
 Start with the full range of possible times.
 Find the middle value.
 Check how many total trips can be done in that time.
 If it meets the requirement, search in the lower half; otherwise, search in the upper half.
 Repeat until you find the minimum time that fulfills the requirement.
Key Steps or Operations:
 Determine the minimum and maximum time values (range).
 Calculate the middle value of the current range.
 Evaluate whether the middle value meets the requirements.
 Narrow the search space accordingly. These steps efficiently guide the search towards the optimal solution.
Algorithmic Patterns:
 Binary Search (reducing the problem size by half at each step)
 Iterative Processing (using a loop to repetitively apply logic) These patterns allow the code to search efficiently and systematically through the sorted data.
The code applies a binary search technique to iteratively narrow down the search space and find the minimum time required for buses to complete the total trips. It leverages logical constructs such as loops and conditions to execute this algorithm, independent of programming language syntax.
Language Agnostic Coding Drills
Let’s dissect the code for the binary search algorithm applied to the problem and categorize them into individual coding concepts or drills.
1. Dissect the Code into Distinct Concepts
 Variable Initialization: Declaring and initializing variables.
 Looping Mechanism: Iterating through elements using a loop.
 Conditional Logic: Implementing ifelse or other conditional constructs.
 Arithmetic Operations: Performing basic arithmetic operations.
 Binary Search Logic: Applying the binary search algorithm.
 Function Definition and Calling: Creating and calling functions.
 Handling Arrays or Lists: Working with data structures like arrays or lists.
 Result Aggregation: Gathering and returning the final result.
2. Order of Increasing Difficulty with Descriptions
 Variable Initialization: Easy  Basic declarations and assignments.
 Arithmetic Operations: Easy  Simple calculations like addition, subtraction.
 Looping Mechanism: Intermediate  Understanding loop structure and control flow.
 Conditional Logic: Intermediate  Implementing decisions based on conditions.
 Handling Arrays or Lists: Intermediate  Requires understanding of data structures.
 Result Aggregation: Intermediate  Aggregating results for final output.
 Function Definition and Calling: Advanced  Writing reusable code blocks.
 Binary Search Logic: Advanced  Applying binary search, needs understanding of above concepts.
3. ProblemSolving Approach from Problem Statement to Final Solution
Understand the Problem: Grasp the core requirement; here, finding the minimum time required for a given number of trips.
Initialize Variables: Set up variables for the minimum and maximum time, and other necessary data.
Implement Looping: Use a loop to iterate through possible time values, applying binary search logic.
Apply Conditional Logic: Inside the loop, use conditions to determine whether to search in the lower or upper half of the current range.
Perform Calculations: Calculate the middle value and other necessary arithmetic.
Work with Lists: Handle any list or array manipulations required for the problem.
Use Binary Search Logic: Embed binary search within the loop to efficiently find the minimum time.
Aggregate Results: Collect the final result and prepare it for output.
Function Creation (if needed): Encapsulate logic in functions for modularity.
The process involves starting from the basic understanding of the problem, initializing variables, applying loops, and embedding binary search logic with necessary calculations. Each of these “drills” builds upon the previous, leading to the final solution. By breaking down the problem into these components, you can create a clear and methodical path to the solution.
Targeted Drills in Python
1. Variable Initialization


2. Arithmetic Operations


3. Looping Mechanism


4. Conditional Logic


5. Handling Arrays or Lists


6. Result Aggregation


7. Function Definition and Calling


8. Binary Search Logic (Problem Specific)


This drill encapsulates the core logic specific to solving the problem using binary search.
Integrating All Drills Together
 Initialize Variables: Set up the min_time and max_time.
 Implement Binary Search: Inside a loop, use arithmetic operations to calculate average_time and use conditional logic to decide the search direction.
 Handle Arrays or Lists: Within the binary search, handle lists or arrays that are specific to the problem. Use functions if necessary.
 Aggregate Results: Once the loop ends, collect and format the result.
Integrating these drills in the order mentioned would build the final solution for the problem. Each drill represents a modular piece of logic, and assembling them in this sequence creates the entire algorithm to solve the initial problem.
Q&A
Similar Problems
Here are 10 distinct problems that use similar underlying concepts:
Find Minimum in Rotated Sorted Array: This problem requires searching for the minimum element in a rotated sorted array. The concept of binary search can be applied here, just like in our original problem.
Search in Rotated Sorted Array: Similar to the original problem, this one involves using binary search in a rotated array to find a specific element. The rotation adds complexity, requiring adapted binary search logic.
Find Peak Element: This problem requires finding a peak element in an unsorted array using binary search. The underlying binary search strategy is what connects it to our original problem.
Capacity To Ship Packages Within D Days: The goal here is to find the minimum capacity of shipment within given days. It’s related to our original problem because it involves binary search to minimize a value based on constraints.
Koko Eating Bananas: This problem is about finding the minimum eating speed to finish all bananas within a certain time, using binary search. The binary search mechanism for finding a minimum value relates it to our original problem.
Split Array Largest Sum: This problem involves splitting an array into parts to minimize the largest sum and utilizes binary search. The minimization aspect and binary search logic relate to our original problem.
Two Sum II  Input array is sorted: This problem involves finding two numbers that add up to a specific target in a sorted array. Binary search or twopointer techniques can be applied, making it related to our original problem.
Find the Duplicate Number: This problem requires finding the duplicate number in an array using the binary search method. It is similar in using binary search to solve an arraybased problem, like our original one.
Find First and Last Position of Element in Sorted Array: This problem involves finding the start and end position of a given target value in a sorted array using binary search, making it related to our original problem.
Search a 2D Matrix II: This problem requires searching a target in a 2D matrix, where the binary search concept can be applied. Its use of binary search for target finding links it to our original problem.