Car Fleet Ii - Leetcode Solution

Problem Statement

You are given two integer arrays of equal length target and position. You have a car that starts at position 0 and can only move to the right. The car cannot move outside the boundaries of the arrays.

The car moves according to the following rules:

For every second, if the car is at position i, it moves to position i + 1 if there is no car at position i + 1. For every second, if the car is at position i, it moves to position i + 1 if there is a car at position i + 1 and the distance between the car at position i + 1 and the car at position i is larger than or equal to the target[i]. Otherwise, the car waits for the car at position i + 1 to move away from position i + 1. Your task is to calculate the minimum time in seconds for the car to reach its destination.

Solution:

In order to calculate the minimum time in seconds for the car to reach its destination, we need to consider each car and calculate the time it will take to reach its target position.

We can start by iterating through each car and keeping track of the maximum time it would take for the car to reach its destination. We can do this by keeping track of the maximum time for each car using a variable maxTime, which we initialize to 0.

For each car, we calculate the time it would take to reach its target position using the formula:

time = (target[i] - target[i-1]) / (position[i] - position[i-1])

Here, i is the index of the current car we are considering and i-1 is the index of the previous car we considered.

If the time is less than or equal to the maximum time we have calculated so far, we add the difference between the maximum time and the time calculated for the current car to the maximum time.

Otherwise, we update the maximum time to be the time calculated for the current car.

Finally, we return the maximum time.

Here is the Python code that implements this solution:

def carFleet(target: int, position: List[int]) -> int: maxTime = 0 n = len(target) cars = sorted(list(zip(target, position)), reverse=True)

for i in range(n):
    time = (cars[i][0] - cars[i-1][0]) / (cars[i][1] - cars[i-1][1]) if i > 0 else 0
    if time <= maxTime:
        maxTime += maxTime - time
    else:
        maxTime = time

return int(maxTime) if maxTime > 0 else n

Time Complexity:

The time complexity of this solution is O(nlogn), where n is the length of the target and position arrays. This is because we sort the cars based on their target distance, which takes O(nlogn). The for loop takes O(n) time. However, the calculation of time for each car takes constant time, so it does not add to the overall time complexity.

Space Complexity:

The space complexity of this solution is O(n), where n is the length of the target and position arrays. This is because we create a list of tuples to represent the cars, which takes O(n) space.

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