LeetCode 857. Minimum Cost to Hire K Workers

Description

https://leetcode.com/problems/minimum-cost-to-hire-k-workers/

There are N workers.  The i-th worker has a quality[i] and a minimum wage expectation wage[i].

Now we want to hire exactly K workers to form a paid group.  When hiring a group of K workers, we must pay them according to the following rules:

1. Every worker in the paid group should be paid in the ratio of their quality compared to other workers in the paid group.
2. Every worker in the paid group must be paid at least their minimum wage expectation.

Return the least amount of money needed to form a paid group satisfying the above conditions.

Example 1:

Input: quality = [10,20,5], wage = [70,50,30], K = 2
Output: 105.00000
Explanation: We pay 70 to 0-th worker and 35 to 2-th worker.

Example 2:

Input: quality = [3,1,10,10,1], wage = [4,8,2,2,7], K = 3
Output: 30.66667
Explanation: We pay 4 to 0-th worker, 13.33333 to 2-th and 3-th workers seperately. 

Note:

1. 1 <= K <= N <= 10000, where N = quality.length = wage.length
2. 1 <= quality[i] <= 10000
3. 1 <= wage[i] <= 10000
4. Answers within 10^-5 of the correct answer will be considered correct.

Explanation

Maintain a min heap of negative quality and the sum of this heap.

For each worker in order of ratio, we know all currently considered workers have lower ratio. We calculate the candidate answer as this ratio times the sum of the smallest K workers in quality.

Python Solution

class Solution:
    def mincostToHireWorkers(self, quality: List[int], wage: List[int], K: int) -> float:
        
        workers = sorted([(w / q, w, q) for w, q in zip(wage, quality)])
        
        heap = []
        
        min_cost = float(inf)
        quality_sum = 0
        
        for ratio, wage, quality in workers:
            heapq.heappush(heap, -quality)
            quality_sum += quality
                        
            if len(heap) > K:
                quality_sum += heapq.heappop(heap)
                
            if len(heap) == K:
                min_cost = min(min_cost, ratio * quality_sum)
                    
        return min_cost

• Time Complexity: ~N(logN)
• Space Complexity: ~N