Problem Informations

• Problem Index: 2594
• Problem Link: Minimum Time to Repair Cars
• Topics: Array, Binary Search

Problem Description

You are given an integer array ranks representing the ranks of some mechanics. $ranks_i$ is the rank of the $i^{th}$ mechanic. A mechanic with a rank r can repair $n$ cars in $r \cdot n^2$ 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.

Example 1:

Input: ranks = [4,2,3,1], cars = 10
Output: 16
Explanation: 
- The first mechanic will repair two cars. The time required is $4 \cdot 2 \cdot 2 = 16$ minutes.
- The second mechanic will repair two cars. The time required is $2 \cdot 2 \cdot 2 = 8$ minutes.
- The third mechanic will repair two cars. The time required is $3 \cdot 2 \cdot 2 = 12$ minutes.
- The fourth mechanic will repair four cars. The time required is $1 \cdot 4 \cdot 4 = 16$ minutes.
It can be proved that the cars cannot be repaired in less than 16 minutes.​​​​​

Example 2:

Input: ranks = [5,1,8], cars = 6
Output: 16
Explanation: 
- The first mechanic will repair one car. The time required is $5 \cdot 1 \cdot 1 = 5$ minutes.
- The second mechanic will repair four cars. The time required is $1 \cdot 4 \cdot 4 = 16$ minutes.
- The third mechanic will repair one car. The time required is $8 \cdot 1 \cdot 1 = 8$ minutes.
It can be proved that the cars cannot be repaired in less than 16 minutes.​​​​​

Constraints:

• $1 \leq ranks.length \leq 10^5$
• $1 \leq ranks[i] \leq 100$
• $1 \leq cars \leq 10^6$

Intuition

The problem involves a set of mechanics, each with a specific rank, and a set number of cars that need to be repaired. The time taken by a mechanic to repair a car is dependent on both the mechanic’s rank and the number of cars they repair. To find the minimal time needed to repair all cars, we need to efficiently allocate car repairs among the mechanics while considering their varying efficiencies due to their ranks. The intuition is that if we can manage the repair tasks such that all cars are repaired within the smallest possible time, then we need to ascertain this minimal time by balancing the load among mechanics optimally.

Approach

The solution utilizes a binary search strategy combined with a helper function that checks if all cars can be repaired within a given time.

1. Initial Setup:

   • First, we determine an initial ‘jump’ size for our binary search. This jump size is calculated as the minimum rank from the list of mechanics multiplied by the square of the total number of cars. This provides an upper-bound estimate on the repair time based on the fastest mechanic working at maximum capacity.

2. Binary Search Strategy:

   • We employ binary search to identify the minimal time required. The binary search operates by progressively halving the ‘jump’ size after verifying the feasibility of repairing all cars within the current time estimate (minTime + jump).
   • We initially set minTime to -1, a value smaller than any possible valid time.
   • The binary search iteratively refines this estimation. In each iteration, we use a helper function to verify whether all cars can be repaired within the current estimated time (minTime + jump). If it’s not possible, this means our current time estimate is too small, and we increase minTime by the current jump size.
   • If it’s possible, the jump size is halved, thus narrowing the range of possible minimal times more precisely.

3. Helper Function:

   • This function checks the feasibility of repairing all cars within a given time t. It calculates how many cars each mechanic can repair by determining the largest integer number of cars, $n$, such that $rank \times n^2 \leq t$. This integer is determined using the formula $n = \sqrt{t / rank}$.
   • For each mechanic, the function subtracts the number of cars they can repair from the total cars. If at any point all cars are accounted for (either exactly or in excess), the function returns true indicating that t is a feasible time. Otherwise, it returns false.

4. Result Calculation:

   • The binary search concludes when the jump size becomes zero, which means we’ve found the minimal time within which all cars can be repaired. Since minTime was increased until the condition was just on the edge of infeasibility, minTime + 1 is returned as the answer, representing the exact minimum time needed for a feasible car repair schedule.

The approach effectively narrows down the solution space using binary search, leveraging the mechanics’ ranks to optimize the car repair distribution and achieving an efficient solution even under the given constraints.

Code

C++

class Solution {
public:
    // Helper function to check if all cars can be repaired within time t
    bool check(const vector<int>& ranks, int cars, long long t) {
        for(int rank : ranks) {
            // Calculate the number of cars that can be repaired by the current mechanic
            cars -= static_cast<long long>(sqrt(t / rank));
            // If all cars are repaired, return true
            if(cars <= 0) return true;
        }
        // If not all cars can be repaired within time t, return false
        return false;
    }

    long long repairCars(vector<int>& ranks, int cars) {
        long long minTime = -1;

        // Determine the initial jump size based on the minimum rank
        long long jump = static_cast<long long>(*min_element(ranks.begin(), ranks.end())) * cars * cars;

        // Use binary search by progressively halving the jump size
        while(jump > 0) {
            // Keep increasing minTime until unable to repair cars within minTime + jump
            while(!check(ranks, cars, minTime + jump)) {
                minTime += jump;
            }
            // Halve the jump size
            jump >>= 1;
        }

        // minTime + 1 is the minimum time required to repair all cars
        return minTime + 1;
    }
};

Python

class Solution:

    # Helper function to check if all cars can be repaired within time t
    def check(self, ranks, cars, t):
        for rank in ranks:
            # Calculate the number of cars that can be repaired by the current mechanic
            cars -= int((t / rank) ** 0.5)
            # If all cars are repaired, return true
            if cars <= 0:
                return True
        # If not all cars can be repaired within time t, return false
        return False

    def repairCars(self, ranks, cars):
        minTime = -1

        # Determine the initial jump size based on the minimum rank
        jump = min(ranks) * cars * cars

        # Use binary search by progressively halving the jump size
        while jump > 0:
            # Keep increasing minTime until unable to repair cars within minTime + jump
            while not self.check(ranks, cars, minTime + jump):
                minTime += jump
            # Halve the jump size
            jump >>= 1

        # minTime + 1 is the minimum time required to repair all cars
        return minTime + 1

Complexity

• Time complexity: The time complexity of the algorithm is $O(n \log(\text{maxRank} \cdot \text{cars}^2))$, where $n$ is the number of mechanics (i.e., the size of the ranks vector) and maxRank is the maximum value in the ranks array. This is because the binary search on the time takes $O(\log(\text{max possible time}))$ iterations, and for each iteration, the check function is invoked which has a complexity of $O(n)$ as it iterates through each mechanic’s rank.

• Space complexity: The space complexity of the algorithm is $O(1)$, as the algorithm uses a constant amount of extra space, regardless of the input size. The computations and checks use primitive data types and do not require additional data structures that scale with input size. The storage for the rank vector itself is not included as it is considered part of the input.