LeetCode 857. Minimum Cost to Hire K Workers DP坑反直觉

本文链接：https://blog.youkuaiyun.com/taoqick/article/details/105623930

本文探讨了一种算法问题，即如何以最小的成本雇佣K名工人，每位工人的支付需按其质量比例分配，并且不低于其最低工资期望。通过分析价格性能比，提出了一种排序并选择最优价格性能比工人的解决方案。

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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:

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.

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

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First, I wrote a DP codes. However, if using dp[i][j] to stand the picked out i workers ending with j, the dp recursive formula is actually not right.

The key points for this problem is to understand the price-performanceratio. Generally, we use performance-price ratio and the customers pursue higher performance-price ratio. As opposed to performance-price ratio, workers prefer higher price-performance ratio. The highest price-performance ratio is suitable for all workers. So we sort by price-perfermance ratio first and pick out K workers with <= some price-perfermance ratio and the least quality.

import heapq

class Solution:
    def mincostToHireWorkers(self, quality, wage, K: int) -> float:
        l = len(quality)
        ratio_arr = sorted([(wage[i] / quality[i], i) for i in range(l)])
        heap, sumq, res = [], 0, float('inf')

        for ratio, idx in ratio_arr:
            sumq += quality[idx]
            heapq.heappush(heap, -quality[idx])
            if (len(heap) > K):
                sumq += heapq.heappop(heap)
            if (len(heap) == K):
                res = min(res, sumq * ratio)
        return res