LeetCode 2594. Minimum Time to Repair Cars Solution in Java, C++, Python & More | Explanation + Code

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 * 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

Solutions

Solution 1: Binary Search

We notice that the longer the repair time, the more cars are repaired. Therefore, we can use the repair time as the target of binary search, and binary search for the minimum repair time.

We define the left and right boundaries of the binary search as left=0, right=ranks[0] × cars × cars. Next, we binary search for the repair time mid, and the number of cars each mechanic can repair is \lfloor √^mid⁄_r \rfloor, where \lfloor x \rfloor represents rounding down. If the number of cars repaired is greater than or equal to cars, it means that the repair time mid is feasible, we reduce the right boundary to mid, otherwise we increase the left boundary to mid+1.

Finally, we return the left boundary.

The time complexity is O(n × log M), and the space complexity is O(1). Here, n is the number of mechanics, and M is the upper bound of the binary search.

PythonJavaC++GoTypeScript

class Solution: def repairCars(self, ranks: List[int], cars: int) -> int: def check(t: int) -> bool: return sum(int(sqrt(t // r)) for r in ranks) >= cars return bisect_left(range(ranks[0] * cars * cars), True, key=check)(code-box)

class Solution { public long repairCars(int[] ranks, int cars) { long left = 0, right = 1L * ranks[0] * cars * cars; while (left < right) { long mid = (left + right) >> 1; long cnt = 0; for (int r : ranks) { cnt += Math.sqrt(mid / r); } if (cnt >= cars) { right = mid; } else { left = mid + 1; } } return left; } }(code-box)

class Solution { public: long long repairCars(vector<int>& ranks, int cars) { long long left = 0, right = 1LL * ranks[0] * cars * cars; while (left < right) { long long mid = (left + right) >> 1; long long cnt = 0; for (int r : ranks) { cnt += sqrt(mid / r); } if (cnt >= cars) { right = mid; } else { left = mid + 1; } } return left; } };(code-box)

func repairCars(ranks []int, cars int) int64 { return int64(sort.Search(ranks[0]*cars*cars, func(t int) bool { cnt := 0 for _, r := range ranks { cnt += int(math.Sqrt(float64(t / r))) } return cnt >= cars })) }(code-box)

function repairCars(ranks: number[], cars: number): number { let left = 0; let right = ranks[0] * cars * cars; while (left < right) { const mid = left + Math.floor((right - left) / 2); let cnt = 0; for (const r of ranks) { cnt += Math.floor(Math.sqrt(mid / r)); } if (cnt >= cars) { right = mid; } else { left = mid + 1; } } return left; }(code-box)