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Minimum Time to Repair Cars

Difficulty: Medium

Problem Description

You are given an integer array ranks representing the ranks of some mechanics. A mechanic with a rank r can repair n cars in r * n² minutes. You are also given an integer cars representing the total number of cars waiting in the garage to be repaired. Return the minimum time taken to repair all the cars. Note: All the mechanics can repair the cars simultaneously.

Key Insights

  • Each mechanic's time to repair cars increases quadratically based on the number of cars they repair.
  • The problem can be framed as finding the minimum time using a binary search approach on the possible time values.
  • The ranks of mechanics determine the efficiency of car repairs, and higher ranks lead to faster repairs for the same number of cars.
  • We can calculate how many cars can be repaired within a certain amount of time by iterating through the mechanics and summing up the maximum cars each can repair within that time.

Space and Time Complexity

Time Complexity: O(n log m), where n is the number of mechanics and m is the maximum time we are searching over.
Space Complexity: O(1), as we are using a constant amount of space for variables.

Solution

The solution utilizes a binary search algorithm to determine the minimum time required to repair all cars. The approach involves:

  1. Setting the bounds for the binary search: the minimum possible time (0) and the maximum possible time based on the highest rank and number of cars.
  2. For each mid-point in our binary search, we calculate how many cars can be repaired by all mechanics within that time.
  3. If the total number of cars that can be repaired is greater than or equal to the required number of cars, we search for a smaller time; otherwise, we search for a larger time.

Code Solutions

def minTimeToRepairCars(ranks, cars):
    def canRepairInTime(time):
        total_cars = 0
        for rank in ranks:
            total_cars += int((time // rank) ** 0.5)  # number of cars this mechanic can repair
        return total_cars >= cars

    left, right = 0, min(ranks) * cars * cars  # upper bound
    while left < right:
        mid = (left + right) // 2
        if canRepairInTime(mid):
            right = mid  # try for a smaller time
        else:
            left = mid + 1  # increase time
    return left
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