<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>

bool cmp(pair<int,double>& a, pair<int,double>& b){
    return a.first < b.first;
}
class Solution {
public:
    double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int k) {
        priority_queue<double> pq;
        vector<pair<int, double>> worker(quality.size());
        for(int i = 0; i < quality.size(); i++){
            double ratio = (double) wage[i]/quality[i];
            worker[i] = {quality[i], ratio};
            pq.push(ratio);
        }
        sort(worker.begin(), worker.end(), cmp);
        double ans = DBL_MAX;
        while(pq.size() >= k){
            double ratio = pq.top(), temp = 0;
            pq.pop();
            int cnt = k, i = 0;
            while(cnt){
                if(worker[i].second <= ratio){
                    cnt--;
                    // cout<<worker[i].first<<" "<<temp<<endl;
                    temp += worker[i].first * ratio;
                    // cout<<ratio<<" "<<worker[i].second<<" "<<temp<<endl;
                }
                i++;
            }
            cout<<endl;
            ans = min(ans, temp);
        }
        return ans;
    }
};
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