What is the time complexity of The Latest Time to Catch a Bus?

The optimal solution has a time complexity of O(n log n + m log m).

What data structures are used to solve The Latest Time to Catch a Bus?

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

What are the key insights for solving The Latest Time to Catch a Bus?

Insight 1: Sorting both `buses` and `passengers` arrays is crucial for efficiently determining which passengers can board which buses based on arrival time. Insight 2: Iterating through the sorted `buses` and keeping track of the number of boarded passengers helps to simulate the boarding process and determine the number of passengers that boarded the last bus.

Which companies ask LeetCode 2332. The Latest Time to Catch a Bus in interviews?

This problem is frequently asked at Zoho.

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The Latest Time to Catch a Bus - LeetCode 2332 Solution

The Latest Time to Catch a Bus - Complete Solution Guide

The Latest Time to Catch a Bus is LeetCode problem 2332, 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 a 0-indexed integer array buses of length n , where buses[i] represents the departure time of the i th bus. You are also given a 0-indexed integer array passengers of length m , where passengers[j] represents the arrival time of the j th passenger. All bus departure times are unique. All passenger arrival times are unique. You are given an integer capacity , which represents the maximum number of passengers that can get on each bus. When a passenger arrives, they will wait in line

Detailed Explanation

The problem describes a scenario where buses depart at specific times (`buses`) and passengers arrive at the bus station at specific times (`passengers`). Each bus has a limited capacity (`capacity`). The goal is to determine the latest possible arrival time at the bus station that guarantees you will be able to catch a bus, without arriving at the same time as any other passenger. Passengers are served in order of their arrival time, and a passenger can board a bus if their arrival time is less than or equal to the bus departure time, and the bus has available capacity.

Solution Approach

The solution simulates the bus boarding process. First, both the `buses` and `passengers` arrays are sorted. Then, the code iterates through the sorted `buses` array. For each bus, it iterates through the `passengers` array, keeping track of the index of the next passenger to board (`p_idx`). If a passenger's arrival time is less than or equal to the bus departure time and the bus has capacity, the passenger is considered to have boarded the bus. The number of passengers boarding the last bus is stored. Finally, the latest possible arrival time is determined based on whether the last bus had free capacity or not. If the last bus had capacity, the latest arrival time is the departure time of the last bus. Otherwise, it's just before the arrival time of the last passenger who boarded. The code then checks if the determined time conflicts with any existing passenger arrival times, decrementing the time if needed to avoid collisions.

Step-by-Step Algorithm

Step 1: Sort the `buses` array in ascending order.
Step 2: Sort the `passengers` array in ascending order.
Step 3: Create a set `passenger_times` to efficiently check if a potential arrival time is already taken by a passenger.
Step 4: Initialize `p_idx` to 0 to keep track of the current passenger being considered.
Step 5: Initialize `boarded_on_last_bus` to 0 to store the number of passengers who boarded the last bus.
Step 6: Iterate through the sorted `buses` array.
Step 7: For each bus, iterate through the `passengers` array from the current `p_idx`, checking if a passenger can board the bus (arrival time <= bus time and bus capacity not full).
Step 8: Update `p_idx` and the number of passengers boarded on the current bus.
Step 9: Update `boarded_on_last_bus` with the number of passengers boarded on the current bus.
Step 10: After processing all buses, determine the `latest_time`. If the last bus had capacity, the `latest_time` is the departure time of the last bus. Otherwise, it's just before the arrival time of the last passenger who boarded.
Step 11: Check if the `latest_time` conflicts with any existing passenger arrival times. If it does, decrement the `latest_time` until it doesn't conflict.
Step 12: Return the `latest_time`.

Key Insights

Insight 1: Sorting both `buses` and `passengers` arrays is crucial for efficiently determining which passengers can board which buses based on arrival time.
Insight 2: Iterating through the sorted `buses` and keeping track of the number of boarded passengers helps to simulate the boarding process and determine the number of passengers that boarded the last bus.
Insight 3: The latest possible arrival time is either the departure time of the last bus if there's still capacity, or just before the arrival time of the last passenger who boarded a bus.

Complexity Analysis

Time Complexity: O(n log n + m log m)

Space Complexity: O(m)

Topics

This problem involves: Array, Two Pointers, Binary Search, Sorting.

Companies

Asked at: Zoho.

The Latest Time to Catch a Bus | LeetCode 2332

Medium

Array Two Pointers Binary Search Sorting

Time: O(n log n + m log m)Space: O(m)

Solution Code

Available in:

import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

class Solution {
    public int latestTimeCatchTheBus(int[] buses, int[] passengers, int capacity) {
        Arrays.sort(buses);
        Arrays.sort(passengers);
        
        Set<Integer> passenger_times = new HashSet<>();
        for (int passenger : passengers) {
            passenger_times.add(passenger);
        }
        
        int p_idx = 0;
        
        int boarded_on_last_bus = 0;
        
        for (int bus_time : buses) {
            int boarded_on_this_bus = 0;
            while (p_idx < passengers.length && passengers[p_idx] <= bus_time && boarded_on_this_bus < capacity) {
                p_idx += 1;
                boarded_on_this_bus += 1;
            }
            boarded_on_last_bus = boarded_on_this_bus;
        }

        int latest_time = 0;
        if (boarded_on_last_bus < capacity) {
            latest_time = buses[buses.length - 1];
        } else {
            latest_time = passengers[p_idx - 1];
        }
        
        while (passenger_times.contains(latest_time)) {
            latest_time -= 1;
        }
        
        return latest_time;
    }
}

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Problem Info

Description

Problem 2332: The Latest Time to Catch a Bus - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search Sorting

Companies

Zoho

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2332

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

12 steps

1
Step 1: Sort the `buses` array in ascending order.
2
Step 2: Sort the `passengers` array in ascending order.
3
Step 3: Create a set `passenger_times` to efficiently check if a potential arrival time is already taken by a passenger.
4
Step 4: Initialize `p_idx` to 0 to keep track of the current passenger being considered.
5
Step 5: Initialize `boarded_on_last_bus` to 0 to store the number of passengers who boarded the last bus.
6
Step 6: Iterate through the sorted `buses` array.
7
Step 7: For each bus, iterate through the `passengers` array from the current `p_idx`, checking if a passenger can board the bus (arrival time <= bus time and bus capacity not full).
8
Step 8: Update `p_idx` and the number of passengers boarded on the current bus.
9
Step 9: Update `boarded_on_last_bus` with the number of passengers boarded on the current bus.
10
Step 10: After processing all buses, determine the `latest_time`. If the last bus had capacity, the `latest_time` is the departure time of the last bus. Otherwise, it's just before the arrival time of the last passenger who boarded.
11
Step 11: Check if the `latest_time` conflicts with any existing passenger arrival times. If it does, decrement the `latest_time` until it doesn't conflict.
12
Step 12: Return the `latest_time`.

💡

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(n log n + m log m)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(m)

📝

Why?

The time complexity is dominated by the sorting algorithms used on both buses and passengers. The loop iterating through buses and passengers is linear. The space complexity stems from the HashSet (or set in Python/C++) used to efficiently check if a time slot is taken by a passenger, requiring space proportional to the number of passengers.

📖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

Binary Search

Instead of linearly checking for each bus and passenger, binary search could be used to find the number of passengers that could board each bus. However, the initial sorting still makes this approach not any more efficient.

Time: O(n log n + m log m)Space: O(m)

Trade-offs: Binary Search adds complexity without improving overall time complexity, as the sorting still dominates.

✨

Key Insights

1Insight 1: Sorting both `buses` and `passengers` arrays is crucial for efficiently determining which passengers can board which buses based on arrival time.
2Insight 2: Iterating through the sorted `buses` and keeping track of the number of boarded passengers helps to simulate the boarding process and determine the number of passengers that boarded the last bus.
3Insight 3: The latest possible arrival time is either the departure time of the last bus if there's still capacity, or just before the arrival time of the last passenger who boarded a bus.

📚

Key Terms Explained

For Students

Array

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `buses` is empty or `passengers` is empty. The code implicitly handles an empty `passengers` case correctly; the loop simply won't run. An empty `buses` array is not explicitly handled and would likely produce incorrect result. Assuming the prompt means the arrays are never empty. If they could be empty the code should have a check for that.
Edge case 2: `capacity` is very large, larger than the number of passengers. The algorithm will still find the latest time, but there will always be capacity on each bus.
Edge case 3: All passengers arrive before the first bus leaves. The algorithm correctly determines the latest time to be less than the first passenger if there's no more room or the time of the first bus if there's room.
Edge case 4: You can arrive at the time of a bus if there's space left. This is implicitly handled in the condition passengers[p_idx] <= bus_time

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the `buses` and `passengers` arrays. This will lead to incorrect boarding order and an incorrect latest arrival time.

✗Common mistake 2: Not accounting for the case where the determined latest arrival time coincides with the arrival time of an existing passenger. This requires decrementing the time until a free slot is found.

✗Common mistake 3: Incorrectly calculating the potential last time. Assuming if the last bus has space, you can simply show up at the time of the last bus. However, you need to make sure this time slot is not taken, or decrement as necessary.

✅

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: Optimizing resource allocation where resources (buses) have limited capacity and requests (passengers) have deadlines (arrival times). For example, assigning tasks to servers with limited processing power.

🏦Real-world scenario 2: Determining the latest safe arrival time at an airport to catch a connecting flight, considering security lines (capacity) and other passengers.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: November 20, 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

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "The Latest Time to Catch a Bus"?

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 two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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