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Topic link and description
https://leetcode.cn/problems/minimum-cost-to-hire-k-workers/
There are n workers. Given two arrays quality and wage, among them, quality[i] represents the work quality of worker i, whose minimum expected wage is wage[i].
Now we want to hire k workers to form a wage group. When hiring a set of k workers, we must pay them wages according to the following rules:
Each worker in a wage group should be paid in proportion to the quality of his work compared to the quality of work of other workers in the same group.
Each worker in a wage group should be paid at least their minimum expected wage.
Given an integer k, return the minimum amount required to form a paying group that satisfies the above criteria. Answers within 10-5 of the actual answer will be accepted. .
Example 1:
Input: quality = [10,20,5], wage = [70,50,30], k = 2
Output: 105.00000
Explanation: We pay worker 0 70 and worker 2 35.
Example 2:
Input: quality = [3,1,10,10,1], wage = [4,8,2,2,7], k = 3
Output: 30.66667
Explanation: We pay worker 0 4, worker 2 and No. 3 pays 13.33333 respectively.
hint:
n == quality.length == wage.length
1 <= k <= n <= 104
1 <= quality[i], wage[i] <= 104
Keywords: Greedy Algorithm Priority Queue
Each worker in a wage group should be paid in proportion to the quality of his work compared to the quality of work of other workers in the same group.
Each worker in a wage group should be paid at least their minimum expected wage.
method one:
run screenshot
the code
public double mincostToHireWorkers(int[] quality, int[] wage, int k) {
// 对工资组中的每名工人，应当按其工作质量与同组其他工人的工作质量的比例来支付工资。
//工资组中的每名工人至少应当得到他们的最低期望工资。
// 首先我们得用公式推导：
// k个人：1-k ,某个人是i
// 对应的质量：q[i],对应的工资期待是w[i]
// 实际的收入是和质量成比例的，也就是假定比例是t[i] = sum(q[1]...q[k])/k
// 那么实际的收入就比例乘自身质量：r[i] = t[i]*q[i]，题目中，必须不低于最低期望，所以r[i] >= w[i]
// 综合可以得到 t[i]*q[i]>=w[i] 等价于 t[i] >= w[i]/q[i]
// 而题目是为了 找k个人，给最少的钱，质量不看重；所以只需要找比例 t[i] 最大的k个人就好了
// 第一步要算出k个人，每人的质量比就好了 w[i]/q[i] > w[j]/q[j]
// 所以才会有w[i]*q[j]-w[j]*q[i] 的排序,还可以避免精度问题
Integer[] sort = IntStream.range(0, quality.length).boxed().toArray(Integer[]::new);
Arrays.sort(sort, (l, r) -> quality[r] * wage[l] - quality[l] * wage[r]);
// 得到了质量比的排序后，有三个变量，质量、花费、最低期望：所以锁定第k人质量最高情况，k-1人质量越低，花费工资越少，然后再对比第k人的质量情况
// 总的结果
double res = 1e9;
// 总的质量
double totalq = 0.0;
// 创建优先级队列
PriorityQueue<Integer> pq = new PriorityQueue<Integer>((a, b) -> b - a);
// 取k个人的质量总和
for (int i = 0; i < sort.length; i++) {
int idx = sort[i];
// 取前k-1个人 性价比高的 得到质量总和
totalq += quality[idx];
// 将当前质量加入优先级队列，质量低的优先
pq.offer(quality[idx]);
if (i >= k - 1) {
// 剩余一个对比花费最少
res = Math.min(res, ((double) wage[idx] / quality[idx]) * totalq);
totalq -= pq.poll();
}
}
return res;
}
end
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