What is the time complexity of Minimum Time to Complete Trips?

The optimal solution has a time complexity of O(nlog(T)).

What data structures are used to solve Minimum Time to Complete Trips?

This problem uses Array, Binary Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Time to Complete Trips?

Insight 1: Binary Search is effective because the problem asks for minimizing the time, and we can determine if a given time is feasible (i.e., enough trips are completed within that time). The search space is monotonic: if a time 't' is enough to complete the trips, then any time greater than 't' is also enough. Insight 2: The number of trips a bus can complete within a given time 't' is simply t // time[i], where time[i] is the time taken by the i-th bus for one trip.

Which companies ask LeetCode 2187. Minimum Time to Complete Trips in interviews?

This problem is frequently asked at Amadeus.

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Minimum Time to Complete Trips - LeetCode 2187 Solution

Minimum Time to Complete Trips - Complete Solution Guide

Minimum Time to Complete Trips is LeetCode problem 2187, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an array time where time[i] denotes the time taken by the i th bus to complete one trip . Each bus can make multiple trips successively ; that is, the next trip can start immediately after completing the current trip. Also, each bus operates independently ; that is, the trips of one bus do not influence the trips of any other bus. You are also given an integer totalTrips , which denotes the number of trips all buses should make in total . Return the minimum time required for all bu

Detailed Explanation

The problem asks us to find the minimum time required for a set of buses to complete a given number of trips. Each bus takes a certain amount of time to complete one trip, and all buses operate independently and simultaneously. The goal is to determine the earliest time at which the total number of trips completed by all buses is at least equal to a target number, `totalTrips`. The input consists of an array `time` representing the time each bus takes for one trip, and an integer `totalTrips` representing the minimum total trips to be completed.

Solution Approach

The solution uses binary search to find the minimum time needed to complete at least `totalTrips`. We start with a search space defined by `low` and `high` values. The `low` value is 1, and the `high` value is initially set to the product of the minimum element in the `time` array and `totalTrips`. In each iteration of the binary search, we calculate the `mid` value, representing a potential time. We then iterate through the `time` array, calculating how many trips each bus can complete in the given `mid` time. We sum these values to get the total number of trips completed. If the total number of trips is greater than or equal to `totalTrips`, we know that the minimum time must be less than or equal to `mid`, so we update `high` to `mid`. Otherwise, we know that the minimum time must be greater than `mid`, so we update `low` to `mid + 1`. This process continues until `low` and `high` converge, at which point `low` will be the minimum time needed to complete at least `totalTrips`.

Step-by-Step Algorithm

Step 1: Initialize `low` to 1 and `high` to `min(time) * totalTrips`. `high` represents the maximum possible time required to complete all trips.
Step 2: Perform binary search while `low < high`.
Step 3: Calculate the `mid` value as `low + (high - low) // 2`.
Step 4: Initialize `trips_count` to 0.
Step 5: Iterate through the `time` array. For each bus with time `t`, calculate the number of trips the bus can complete as `mid // t` and add it to `trips_count`.
Step 6: If `trips_count >= totalTrips`, update `high` to `mid` because the current `mid` value is a possible solution and we want to find the minimum possible time.
Step 7: Otherwise, if `trips_count < totalTrips`, update `low` to `mid + 1` because the current `mid` value is not enough to complete the required trips.
Step 8: After the binary search loop finishes, return `low`, which represents the minimum time needed to complete at least `totalTrips`.

Key Insights

Insight 1: Binary Search is effective because the problem asks for minimizing the time, and we can determine if a given time is feasible (i.e., enough trips are completed within that time). The search space is monotonic: if a time 't' is enough to complete the trips, then any time greater than 't' is also enough.
Insight 2: The number of trips a bus can complete within a given time 't' is simply t // time[i], where time[i] is the time taken by the i-th bus for one trip.
Insight 3: The search space for time is bounded. The lower bound is 1 (minimum possible time), and the upper bound can be the minimum time taken by any bus multiplied by the total number of trips required (min(time) * totalTrips). This gives us a reasonable range for binary search to work.

Complexity Analysis

Time Complexity: O(nlog(T))

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search.

Companies

Asked at: Amadeus.

Minimum Time to Complete Trips | LeetCode 2187

Medium

Array Binary Search

Time: O(nlog(T))Space: O(1)

Solution Code

Available in:

class Solution {
    public long minimumTime(int[] time, int totalTrips) {
        long low = 1;
        long high = (long) time[0] * totalTrips;
        for (int t : time) {
            high = Math.min(high, (long) t * totalTrips);
        }


        while (low < high) {
            long mid = low + (high - low) / 2;
            
            long trips_count = 0;
            for (int t : time) {
                trips_count += mid / t;
                if (trips_count >= totalTrips) {
                    break;
                }
            }
            
            if (trips_count >= totalTrips) {
                high = mid;
            } else {
                low = mid + 1;
            }
        
        }
        return low;
    }
}

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Problem Info

Description

Problem 2187: Minimum Time to Complete Trips - Detailed explanation and solution approaches.

Topics

Array Binary Search

Companies

Amadeus

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2187

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize `low` to 1 and `high` to `min(time) * totalTrips`. `high` represents the maximum possible time required to complete all trips.
2
Step 2: Perform binary search while `low < high`.
3
Step 3: Calculate the `mid` value as `low + (high - low) // 2`.
4
Step 4: Initialize `trips_count` to 0.
5
Step 5: Iterate through the `time` array. For each bus with time `t`, calculate the number of trips the bus can complete as `mid // t` and add it to `trips_count`.
6
Step 6: If `trips_count >= totalTrips`, update `high` to `mid` because the current `mid` value is a possible solution and we want to find the minimum possible time.
7
Step 7: Otherwise, if `trips_count < totalTrips`, update `low` to `mid + 1` because the current `mid` value is not enough to complete the required trips.
8
Step 8: After the binary search loop finishes, return `low`, which represents the minimum time needed to complete at least `totalTrips`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(nlog(T))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the binary search which runs for O(log(T)) iterations. Inside each iteration, we iterate through the array `time` to calculate `trips_count`, taking O(n) time. Therefore, the overall time complexity is O(n log(T)). The space complexity is O(1) because we only use a few variables to store intermediate results, which do not depend on the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Linear Search

Iterate through possible time values starting from 1, and for each time value calculate the total number of trips completed. Stop when the total number of trips is greater than or equal to `totalTrips`. This approach would not be efficient.

Time: O(T * n), where T is the minimum time to complete all trips and n is the length of the `time` array.Space: O(1)

Trade-offs: Linear search is much slower than binary search because it iterates through all possible time values until the condition is met. Use the main solution, binary search, for a faster solution.

✨

Key Insights

1Insight 1: Binary Search is effective because the problem asks for minimizing the time, and we can determine if a given time is feasible (i.e., enough trips are completed within that time). The search space is monotonic: if a time 't' is enough to complete the trips, then any time greater than 't' is also enough.
2Insight 2: The number of trips a bus can complete within a given time 't' is simply t // time[i], where time[i] is the time taken by the i-th bus for one trip.
3Insight 3: The search space for time is bounded. The lower bound is 1 (minimum possible time), and the upper bound can be the minimum time taken by any bus multiplied by the total number of trips required (min(time) * totalTrips). This gives us a reasonable range for binary search to work.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: If `time` contains only one element, the binary search will still converge to the correct result. The upper bound will be `time[0] * totalTrips`, and the binary search will effectively find the minimum time.
Edge case 2: If `totalTrips` is 1 and `time` contains only one element, the algorithm will correctly return the value of that element. The binary search will quickly converge to the time taken by the single bus.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow when calculating `min(time) * totalTrips`. This is handled by using `long` data types for `high` and other relevant variables, especially in Java and C++.

✗Common mistake 2: Incorrectly implementing the binary search. Specifically, not handling the cases where `low` and `high` are adjacent and not correctly updating `low` and `high` based on whether `trips_count` is greater than or equal to `totalTrips`.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Task scheduling where multiple workers (buses) are available, each with different processing speeds (time), and the goal is to find the minimum time to complete a certain number of tasks (trips).

🏦Real-world scenario 2: Resource allocation where multiple resources (buses) are available, each with different efficiencies (time), and the goal is to find the minimum time to meet a certain demand (trips).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 28, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Time to Complete Trips"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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