LeetCode 2594 – Minimum Time To Repair Cars

Although this problem is not frequently asked by companies, it offers an interesting perspective. Initially, this problem may seem to require a complex solution, but it can be efficiently solved using binary search.

Binary Search Approach

The key is to frame the problem as finding the minimum time required to repair all cars, within a range from 1 to the maximum possible time. The maximum possible time is determined by the minimum rank multiplied by the square of the number of cars. Since the number of cars is always greater than 0, the theoretical minimum possible time is 1, and the solution lies within this range.

Therefore, the problem is transformed into a binary search, where at each iteration, we check if all cars can be repaired within a given time. If yes, we search the left half for a potentially lower time; otherwise, we search the right half.

We know that a mechanic with rank ‘r’ takes time to repair ‘n’ cars, so within time ‘t’, they can repair​ sqrt(t/r) cars. We need to create a helper function that returns a boolean indicating whether all cars can be repaired within a given time ‘t’. We have to take the floor and then we sum the values to determine number of cars that can be repaired in time t. If this value is greater than or equal to the desired number of cars, we will return True.

def can_repair(self, t,n,ranks):
        n_each_mech = [int((t/r)**0.5) for r in ranks]
        return sum(n_each_mech)>=n

The solution to the problem.

class Solution:
    def repairCars(self, ranks: List[int], cars: int) -> int:
        min_rank = min(ranks)
        low = 1
        high = min_rank*cars*cars

        min_time = None

        while low <= high:
            mid = (high+low)//2

            if self.can_repair(mid,cars,ranks):
                min_time = mid
                high = mid-1

            else:
                low = mid+1
        
        return min_time

    def can_repair(self, t,n,ranks):
        n_each_mech = [int((t/r)**0.5) for r in ranks]
        return sum(n_each_mech)>=n

Complexity Analysis

The time complexity is O(N log(C)), where N is the number of mechanics and C is the range of possible times. The binary search contributes a factor of O(log(C)), and the can_repair function contributes a factor of O(N). The space complexity is O(1) as we are only using constant extra space.

Let me know if you would like to cover any additional aspects or if you have any further questions.