N workers. The
i-th worker has a
quality[i] and a minimum wage expectation
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:
- Every worker in the paid group should be paid in the ratio of their quality compared to other workers in the paid group.
- 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.
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.
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.
1 <= K <= N <= 10000, where
N = quality.length = wage.length
1 <= quality[i] <= 10000
1 <= wage[i] <= 10000
- Answers within
10^-5of the correct answer will be considered correct.
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.
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