What is the time complexity of Car Pooling?

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

What data structures are used to solve Car Pooling?

This problem uses Array, Sorting, Heap (Priority Queue), Simulation, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Car Pooling?

Insight 1: The problem can be modeled as a series of passenger changes at specific locations. Increasing the passenger count at the pick-up location and decreasing it at the drop-off location. Insight 2: Using a timeline (array) to track the net change in passenger count at each location allows us to efficiently simulate the car's occupancy as it moves along the route.

Which companies ask LeetCode 1094. Car Pooling in interviews?

This problem is frequently asked at Flipkart, Lyft, Zepto.

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Car Pooling - LeetCode 1094 Solution

Car Pooling - Complete Solution Guide

Car Pooling is LeetCode problem 1094, 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

There is a car with capacity empty seats. The vehicle only drives east (i.e., it cannot turn around and drive west). You are given the integer capacity and an array trips where trips[i] = [numPassengers i , from i , to i ] indicates that the i th trip has numPassengers i passengers and the locations to pick them up and drop them off are from i and to i respectively. The locations are given as the number of kilometers due east from the car's initial location. Return true if it is possible to pick

Detailed Explanation

The problem describes a car pooling scenario where a car with a limited capacity travels east, picking up and dropping off passengers at various locations. The input consists of a list of trips, where each trip is defined by the number of passengers, the pick-up location (kilometers from the starting point), and the drop-off location. The problem asks us to determine if it's possible to complete all the trips without exceeding the car's capacity at any point along the route. The locations are integers between 0 and 1000, inclusive, and the capacity is an integer greater than or equal to 1. We must return true if all trips can be completed, and false otherwise.

Solution Approach

The provided solution uses a timeline array to simulate the passenger changes at each location. For each trip, the number of passengers is added to the timeline at the pick-up location and subtracted from the timeline at the drop-off location. Then the solution iterates through the timeline, calculating the current number of passengers in the car. If at any point the current number of passengers exceeds the car's capacity, the solution returns false. Otherwise, after iterating through the entire timeline, the solution returns true.

Step-by-Step Algorithm

Step 1: Initialize a timeline array of size 1001 (since locations are from 0 to 1000) with all elements set to 0.
Step 2: Iterate through the trips array. For each trip [numPassengers, fromLocation, toLocation], add numPassengers to timeline[fromLocation] and subtract numPassengers from timeline[toLocation].
Step 3: Initialize currentPassengers to 0.
Step 4: Iterate through the timeline array. For each location, add the value in timeline[location] to currentPassengers.
Step 5: If currentPassengers exceeds capacity at any point, return false.
Step 6: After iterating through the entire timeline, return true.

Key Insights

Insight 1: The problem can be modeled as a series of passenger changes at specific locations. Increasing the passenger count at the pick-up location and decreasing it at the drop-off location.
Insight 2: Using a timeline (array) to track the net change in passenger count at each location allows us to efficiently simulate the car's occupancy as it moves along the route.
Insight 3: Iterating through the timeline and accumulating the passenger changes allows us to check if the capacity is ever exceeded. If it exceeds the capacity at any point, the problem has no solution.

Complexity Analysis

Time Complexity: O(N)

Space Complexity: O(1)

Topics

This problem involves: Array, Sorting, Heap (Priority Queue), Simulation, Prefix Sum.

Companies

Asked at: Flipkart, Lyft, Zepto.

Car Pooling | LeetCode 1094

Medium

Array Sorting Heap (Priority Queue)Simulation Prefix Sum

Time: O(N)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean carPooling(int[][] trips, int capacity) {
        int[] timeline = new int[1001];

        for (int[] trip : trips) {
            int numPassengers = trip[0];
            int startLocation = trip[1];
            int endLocation = trip[2];

            timeline[startLocation] += numPassengers;
            timeline[endLocation] -= numPassengers;
        }

        int currentPassengers = 0;
        for (int passengerChange : timeline) {
            currentPassengers += passengerChange;
            if (currentPassengers > capacity) {
                return false;
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 1094: Car Pooling - Detailed explanation and solution approaches.

Topics

Array Sorting Heap (Priority Queue)Simulation Prefix Sum

Companies

FlipkartLyftZepto

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:1094

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

6 steps

1
Step 1: Initialize a timeline array of size 1001 (since locations are from 0 to 1000) with all elements set to 0.
2
Step 2: Iterate through the trips array. For each trip [numPassengers, fromLocation, toLocation], add numPassengers to timeline[fromLocation] and subtract numPassengers from timeline[toLocation].
3
Step 3: Initialize currentPassengers to 0.
4
Step 4: Iterate through the timeline array. For each location, add the value in timeline[location] to currentPassengers.
5
Step 5: If currentPassengers exceeds capacity at any point, return false.
6
Step 6: After iterating through the entire timeline, return true.

💡

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)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the iteration through the timeline array, which has a fixed size. The space complexity is determined by the timeline array, which has a fixed size, so it's O(1).

📖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

Sorting and Priority Queue

Sort the trips based on the start location. Then, use a priority queue to track the end locations of current passengers. At each start location, add the number of passengers. Remove passengers from the priority queue when their drop off location is reached. Compare the current passenger load with the capacity.

Time: O(T log T), where T is the number of trips. Sorting takes O(T log T), and priority queue operations take O(log T).Space: O(T) in the worst case, as the priority queue could potentially hold all trips.

Trade-offs: This approach might be preferred if the range of locations is much larger than the number of trips, as it avoids creating a large timeline array. However, the sorting overhead and priority queue operations make it less efficient when the range is small and the number of trips is large.

✨

Key Insights

1Insight 1: The problem can be modeled as a series of passenger changes at specific locations. Increasing the passenger count at the pick-up location and decreasing it at the drop-off location.
2Insight 2: Using a timeline (array) to track the net change in passenger count at each location allows us to efficiently simulate the car's occupancy as it moves along the route.
3Insight 3: Iterating through the timeline and accumulating the passenger changes allows us to check if the capacity is ever exceeded. If it exceeds the capacity at any point, the problem has no solution.

📚

Key Terms Explained

For Students

Array

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

Sorting

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

Edge Cases

Edge case 1: Trips that start and end at the same location. The algorithm correctly handles this case by adding and then subtracting the number of passengers at that location, resulting in no net change.
Edge case 2: Overlapping trips. The algorithm correctly handles overlapping trips by accumulating the passenger changes at each location. This allows it to accurately track the current number of passengers in the car at each point along the route.
Edge case 3: Empty trip list. If trips is empty, the algorithm correctly returns true as there are no passengers to pick up or drop off.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to subtract the number of passengers at the drop-off location. This would lead to an inaccurate calculation of the current number of passengers and potentially a false negative result.

✗Common mistake 2: Not considering overlapping trips. If the algorithm doesn't correctly accumulate the passenger changes at each location, it may incorrectly determine that the car's capacity is exceeded.

✗Common mistake 3: Using wrong data type for the number of passengers. If trips[i][0] is larger than the capacity, it could lead to integer overflow and cause unexpected behavior.

✅

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: Resource allocation. The algorithm can be used to determine if a limited resource (e.g., bandwidth, server capacity, warehouse space) is sufficient to meet the demands of various requests at different times.

🏦Real-world scenario 2: Scheduling and capacity planning. This pattern can be used in scenarios such as scheduling deliveries to stores from a warehouse or scheduling appointments for a service, ensuring resources are not over-allocated.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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 15, 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

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

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

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.

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 "Car Pooling"?

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

🤖 Need help with this problem?