857. 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.

根据rule1，工资应该与工作质量成正比，所以w[i]/q[i] = w[j]/q[j]。可以求出每个员工的w[i]/q[i] ratio，并排序。维护一个max heap，按quality排序。按ratio从小到大的顺序，依次向heap中添加quality，并累计quality sum。当heap的size大于K时，应该poll quality最高的员工（因为总工资 = current ratio * quality sum）。poll完，heap的size等于K时，记录一下此时的cost，并与前一步的cost比较，取最小值。遍历完全部ratio后，min cost就是答案。

注意：ratio的计算和处理，double的位置

时间：O(NlogN)+O(NlogK) 排序和堆

class Solution {
    public double mincostToHireWorkers(int[] quality, int[] wage, int K) {
        double[][] worker = new double[wage.length][2];
        for(int i = 0; i < wage.length; i++) {
            worker[i][0] = (double) wage[i] / quality[i];
            worker[i][1] = quality[i];
        }
        Arrays.sort(worker, (a, b) -> Double.compare(a[0], b[0]));
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b) -> b - a);
        double min_cost = Integer.MAX_VALUE;
        int qsum = 0;
        for(double[] w : worker) {
            maxHeap.add((int)w[1]);
            qsum += w[1];
            if(maxHeap.size() > K) {
                int tmp = maxHeap.poll();
                qsum -= tmp;
            }
            if(maxHeap.size() == K) {
                min_cost = Math.min(min_cost, w[0] * qsum);
            }
        }
        return min_cost;
    }
}