<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) {


vector<pair<double,double>> ratio_quality ; 
int n = quality.size() ;

for ( int i = 0 ; i<n ; i++) {
double curr_ratio = double(double(wage[i])/double(quality[i])) ; 
//cout<<curr_ratio ;
ratio_quality.push_back({curr_ratio,quality[i]}) ; 

}
sort(ratio_quality.begin(),ratio_quality.end()) ; 
multiset<double> current_qualities ; 
double sum_of_qualities = 0 ; 
double ans = 100000000000 ; 

for ( auto &h:ratio_quality) {
if (current_qualities.size()<k) {
current_qualities.insert(h.second) ; 
sum_of_qualities = sum_of_qualities + h.second ; 
}
else {
if (h.second<*current_qualities.rbegin() ) {
sum_of_qualities = sum_of_qualities - *current_qualities.rbegin() ;
current_qualities.erase(current_qualities.find(*current_qualities.rbegin())) ; 
current_qualities.insert(h.second) ; 
sum_of_qualities  = sum_of_qualities + h.second ;

}
}
if (current_qualities.size()==k) {
double val = sum_of_qualities*h.first ; 
//cout<<val<<" "<<ans<<" " ;
if (val<ans) {
ans = val ;
}
}
}
return ans ;


}
};


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