What is the time complexity of Car Fleet?

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

What data structures are used to solve Car Fleet?

This problem uses Array, Stack, Sorting, Monotonic Stack. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Car Fleet?

Insight 1: Sorting the cars by their starting positions in reverse order simplifies the fleet detection process. By processing cars from closest to the target to farthest, we can easily determine if a car will join an existing fleet. Insight 2: The key factor determining fleet formation is the time it takes for each car to reach the target. If a car's arrival time is less than or equal to the current fleet's arrival time, it will join that fleet.

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Car Fleet - LeetCode 853 Solution

Q: Which companies ask LeetCode 853. Car Fleet in interviews?

This problem is frequently asked at BNY Mellon, BharatPe, GE Healthcare, Infosys, Nutanix and 1 more companies.

Car Fleet - Complete Solution Guide

Car Fleet is LeetCode problem 853, 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 are n cars at given miles away from the starting mile 0, traveling to reach the mile target . You are given two integer arrays position and speed , both of length n , where position[i] is the starting mile of the i th car and speed[i] is the speed of the i th car in miles per hour. A car cannot pass another car, but it can catch up and then travel next to it at the speed of the slower car. A car fleet is a single car or a group of cars driving next to each other. The speed of the car fleet

Detailed Explanation

The problem describes a scenario with multiple cars moving towards a target destination. Each car has a starting position and speed. Cars can form fleets, where a car catches up to another and they travel together at the speed of the slower car. The goal is to determine the number of car fleets that will reach the target.

Solution Approach

The provided solution sorts the cars by their initial positions in descending order. It then iterates through the sorted cars, calculating the arrival time for each car to reach the target. If a car's arrival time is greater than the current fleet's arrival time, it means this car forms a new fleet. Otherwise, it joins the existing fleet. The solution maintains a count of the number of fleets formed.

Step-by-Step Algorithm

Step 1: Create a list of car objects (or pairs) containing position and speed for each car.
Step 2: Sort the list of cars in descending order based on their position. Cars closer to the target come first.
Step 3: Initialize a variable `fleets` to 0, to store the number of fleets.
Step 4: Initialize a variable `current_fleet_arrival_time` to 0.0, representing the arrival time of the latest formed fleet.
Step 5: Iterate through the sorted list of cars.
Step 6: For each car, calculate its arrival time as `(target - position) / speed`.
Step 7: Compare the car's arrival time with `current_fleet_arrival_time`.
Step 8: If the car's arrival time is greater than `current_fleet_arrival_time`, it means the car will not join the current fleet, so increment `fleets` by 1 and update `current_fleet_arrival_time` to the car's arrival time.
Step 9: Return the final `fleets` count.

Key Insights

Insight 1: Sorting the cars by their starting positions in reverse order simplifies the fleet detection process. By processing cars from closest to the target to farthest, we can easily determine if a car will join an existing fleet.
Insight 2: The key factor determining fleet formation is the time it takes for each car to reach the target. If a car's arrival time is less than or equal to the current fleet's arrival time, it will join that fleet.
Insight 3: Since cars cannot pass each other, we only need to track the arrival time of the slowest car (the one that arrives latest) in the current fleet. A new fleet is formed only when the current car takes longer to reach the target than the current fleet's arrival time.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Stack, Sorting, Monotonic Stack.

Companies

Asked at: BNY Mellon, BharatPe, GE Healthcare, Infosys, Nutanix, Snap.

Car Fleet | LeetCode 853

Medium

Array Stack Sorting Monotonic Stack

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.Comparator;

class Solution {
    public int carFleet(int target, int[] position, int[] speed) {
        int n = position.length;
        double[][] cars = new double[n][2];
        
        for (int i = 0; i < n; i++) {
            cars[i][0] = position[i];
            cars[i][1] = speed[i];
        }
        
        Arrays.sort(cars, Comparator.comparingDouble(a -> -a[0]));
        
        int fleets = 0;
        double current_fleet_arrival_time = 0.0;
        
        for (double[] car : cars) {
            double pos = car[0];
            double spd = car[1];
            
            double arrival_time = (target - pos) / spd;
            
            if (arrival_time > current_fleet_arrival_time) {
                fleets++;
                current_fleet_arrival_time = arrival_time;
            }
        }
        
        return fleets;
    }
}

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Problem Info

Description

Problem 853: Car Fleet - Detailed explanation and solution approaches.

Topics

Array Stack Sorting Monotonic Stack

Companies

BNY MellonBharatPeGE HealthcareInfosysNutanixSnap

Stats

Difficulty:Medium

Languages:4

Companies:6

Problem #:853

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

9 steps

1
Step 1: Create a list of car objects (or pairs) containing position and speed for each car.
2
Step 2: Sort the list of cars in descending order based on their position. Cars closer to the target come first.
3
Step 3: Initialize a variable `fleets` to 0, to store the number of fleets.
4
Step 4: Initialize a variable `current_fleet_arrival_time` to 0.0, representing the arrival time of the latest formed fleet.
5
Step 5: Iterate through the sorted list of cars.
6
Step 6: For each car, calculate its arrival time as `(target - position) / speed`.
7
Step 7: Compare the car's arrival time with `current_fleet_arrival_time`.
8
Step 8: If the car's arrival time is greater than `current_fleet_arrival_time`, it means the car will not join the current fleet, so increment `fleets` by 1 and update `current_fleet_arrival_time` to the car's arrival time.
9
Step 9: Return the final `fleets` count.

💡

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)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The `sort` function in Python, Java and C++ has an average time complexity of O(n log n). The rest of the algorithm iterates through the sorted list once, which takes O(n) time. Therefore, the overall time complexity is O(n log n). The space complexity arises from storing the car information (position and speed), which requires O(n) space.

📖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

Without Sorting (less efficient)

An approach that avoids sorting could involve comparing each car with every other car to detect fleet formation. However, this would be much less efficient.

Time: O(n^2)Space: O(1) or O(n) depending on implementation details. O(1) if only count is needed, O(n) if keeping track of formed fleets.

Trade-offs: This approach is less efficient than the sorting-based solution and should be avoided unless memory constraints are extremely tight and the number of cars is very small.

✨

Key Insights

1Insight 1: Sorting the cars by their starting positions in reverse order simplifies the fleet detection process. By processing cars from closest to the target to farthest, we can easily determine if a car will join an existing fleet.
2Insight 2: The key factor determining fleet formation is the time it takes for each car to reach the target. If a car's arrival time is less than or equal to the current fleet's arrival time, it will join that fleet.
3Insight 3: Since cars cannot pass each other, we only need to track the arrival time of the slowest car (the one that arrives latest) in the current fleet. A new fleet is formed only when the current car takes longer to reach the target than the current fleet's arrival time.

📚

Key Terms Explained

For Students

Array

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

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Sorting

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

Edge Cases

Edge case 1: When there is only one car. The solution correctly identifies this as a single fleet.
Edge case 2: When all cars form a single fleet. The algorithm correctly calculates the arrival times and determines that only one fleet exists.
Edge case 3: When the input arrays `position` or `speed` are empty. The provided solutions handle this implicitly because the loops will not execute, and the initialized `fleets` value (which is 0) will be returned.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the cars by their position. If cars are not processed in order from closest to farthest, it becomes difficult to accurately determine fleet formation.

✗Common mistake 2: Incorrectly calculating the arrival time. Ensure the correct formula `(target - position) / speed` is used. Integer division errors can also occur if not using floating-point numbers for the calculation.

✅

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: Traffic simulation. Determining the number of distinct groups of vehicles traveling along a highway.

🏦Real-world scenario 2: Optimizing delivery routes. Analyzing a fleet of delivery trucks to determine how many separate delivery groups will arrive at a central hub.

💡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: October 12, 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

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

@CodeNinjaJul 14

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

@TechLeadJul 7

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

@DevExpert92Jun 30

❓

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.

What is a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

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 "Car Fleet"?

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

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