2665. Minimum Time To Repair Cars¶

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2665. Minimum Time to Repair Cars

Medium

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 * 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.

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 * 2 * 2 = 16 minutes.
- The second mechanic will repair two cars. The time required is 2 * 2 * 2 = 8 minutes.
- The third mechanic will repair two cars. The time required is 3 * 2 * 2 = 12 minutes.
- The fourth mechanic will repair four cars. The time required is 1 * 4 * 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 * 1 * 1 = 5 minutes.
- The second mechanic will repair four cars. The time required is 1 * 4 * 4 = 16 minutes.
- The third mechanic will repair one car. The time required is 8 * 1 * 1 = 8 minutes.
It can be proved that the cars cannot be repaired in less than 16 minutes.​​​​​

Constraints:

• 1 <= ranks.length <= 105
• 1 <= ranks[i] <= 100
• 1 <= cars <= 106

Solution¶

class Solution {
    public long repairCars(int[] ranks, int cars) {
        int n = ranks.length;
        long ans = -1, low = 1, high = (long)(1e15);
        while (low <= high) {
            long mid = low + (high - low) / 2;
            if (ok(mid, ranks, cars)) {
                ans = mid;
                high = mid - 1;
            }
            else low = mid + 1;
        }
        return ans;
    }
    private boolean ok(long mid, int ranks[] , int cars) {
        int n = ranks.length;
        long count = 1, total = 0;
        for (int i = 0; i < n; i++) {
            long current = ranks[i];
            total += (long)(Math.sqrt(mid / current));
        }
        return total >= cars;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here