<p>There are <code>n</code> workers. You are given two integer arrays <code>quality</code> and <code>wage</code> where <code>quality[i]</code> is the quality of the <code>i<sup>th</sup></code> worker and <code>wage[i]</code> is the minimum wage expectation for the <code>i<sup>th</sup></code> worker.</p>

<p>We want to hire exactly <code>k</code> workers to form a <strong>paid group</strong>. To hire a group of <code>k</code> workers, we must pay them according to the following rules:</p>

<ol>
	<li>Every worker in the paid group must be paid at least their minimum wage expectation.</li>
	<li>In the group, each worker&#39;s pay must be directly proportional to their quality. This means if a worker&rsquo;s quality is double that of another worker in the group, then they must be paid twice as much as the other worker.</li>
</ol>

<p>Given the integer <code>k</code>, return <em>the least amount of money needed to form a paid group satisfying the above conditions</em>. Answers within <code>10<sup>-5</sup></code> of the actual answer will be accepted.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> quality = [10,20,5], wage = [70,50,30], k = 2
<strong>Output:</strong> 105.00000
<strong>Explanation:</strong> We pay 70 to 0<sup>th</sup> worker and 35 to 2<sup>nd</sup> worker.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> quality = [3,1,10,10,1], wage = [4,8,2,2,7], k = 3
<strong>Output:</strong> 30.66667
<strong>Explanation:</strong> We pay 4 to 0<sup>th</sup> worker, 13.33333 to 2<sup>nd</sup> and 3<sup>rd</sup> workers separately.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == quality.length == wage.length</code></li>
	<li><code>1 &lt;= k &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= quality[i], wage[i] &lt;= 10<sup>4</sup></code></li>
</ul>

#include <bits/stdc++.h>

using namespace std;

class Solution {
public:
    double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int k) {
        int n = quality.size();

        vector<int> rank(n);
        for (int i = 0; i < n; i++) rank[i] = i;
        sort(rank.begin(), rank.end(), [&quality, &wage](int a, int b){
            return wage[a] * quality[b] < wage[b] * quality[a];
        });


        double result = numeric_limits<double>::max();

        int sum_quality = 0;
        priority_queue<int> max_heap;
        for(int i = 0; i < n; i++) {
            sum_quality += quality[rank[i]];
            max_heap.push(quality[rank[i]]);

            if (max_heap.size() > k) {
                sum_quality -= max_heap.top();
                max_heap.pop();
            }

            if (max_heap.size() == k) {
                result = min(result, sum_quality * (static_cast<double>(wage[rank[i]]) / static_cast<double>(quality[rank[i]])));
            }
        }

        return result;
    }
};
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