<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>

class Solution {
public:
    double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int k) {
        // base salary * qi/qmin >= wi
        // base salary >= wi * qmin / qi
        // base salary = max(wi * qmin / qi)

        // salary_i = base salary * qi / qmin
        // salary_i = max(wi * qmin / qi) * qi / qmin 
        // salary_i = max(wi / qi) * qi

        // total salary = max(wi / qi) * sum(qi)

        int N = wage.size();
        vector<int> idxs(N);
        
        iota(idxs.begin(), idxs.end(), 0);

        sort(idxs.begin(), idxs.end(), [&] (int i1, int i2) {
            double ratio1 = (double) wage[i1] / quality[i1];
            double ratio2 = (double) wage[i2] / quality[i2];
            return ratio1 < ratio2;
        });

        priority_queue<int> pq;
        int sum_qi = 0;
        double ans = 1e9;
        for (auto i : idxs) {
            double ratio = (double) wage[i] / quality[i];
            sum_qi += quality[i];

            if (pq.size() == k) {
                sum_qi -= pq.top();
                pq.pop();
            }

            pq.push(quality[i]);

            if (pq.size() == k) {
                ans = min(ans, ratio * sum_qi);
            }
        }
       
        return ans;
    }
};


// quality = [1, 10, 100]
// wage    = [1, 20, 1]

// all employed
// cost = 1 + 20 + 200

// consider K cheapest workers.
// the only way to potentially decrease the cost
// is to replace the lowest quality worker with sb of higher quality.

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