How do you solve Car Fleet on LeetCode?

Car Fleet (LeetCode #853) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Cars that reach the target at the same time form a fleet.

What is the brute force solution for Car Fleet?

The brute force approach for Car Fleet is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each car against every other car to see if they can form a fleet. This method is straightforward but inefficient as it involves comparing each car's position and speed. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Car Fleet?

Car Fleet uses the following concepts: Array, Stack, Sorting, Monotonic Stack. The recommended patterns to study are: Sorting, Greedy Algorithms.

What are the interview tips for Car Fleet?

Explain your thought process clearly while coding. Consider edge cases, such as when there's only one car. Practice explaining your solution as you write code.

What are common mistakes when solving Car Fleet?

Not sorting the cars before processing their times. Confusing the time to reach the target with the position of the cars.

#853

Car Fleet

Medium

ArrayStackSortingMonotonic StackSortingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

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

The optimal approach sorts the cars by their starting positions and calculates the time to reach the target. By iterating from the back, we can efficiently count fleets based on the time taken to reach the target.

⚙️

Algorithm

3 steps

1Step 1: Pair each car's position with its time to reach the target and sort them by position in descending order.
2Step 2: Initialize a counter for fleets and a variable to track the last time taken to reach the target.
3Step 3: Iterate through the sorted list and compare each car's time with the last recorded time. If it's greater, it forms a new fleet.

solution.py16 lines

1# Full working Python code
2
3def car_fleet(target, position, speed):
4    times = [(target - p) / s for p, s in zip(position, speed)]
5    cars = sorted(zip(position, times), reverse=True)
6    fleets = 0
7    last_time = 0
8
9    for pos, time in cars:
10        if time > last_time:
11            fleets += 1
12            last_time = time
13    return fleets
14
15# Example usage
16print(car_fleet(12, [10, 8, 0, 5, 3], [2, 4, 1, 1, 3]))

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) because we store the times in an array.

1Cars that reach the target at the same time form a fleet.
2Sorting by position allows us to efficiently determine fleet formation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.