<p>You are given a sorted integer array <code>arr</code> containing <code>1</code> and <strong>prime</strong> numbers, where all the integers of <code>arr</code> are unique. You are also given an integer <code>k</code>.</p>

<p>For every <code>i</code> and <code>j</code> where <code>0 &lt;= i &lt; j &lt; arr.length</code>, we consider the fraction <code>arr[i] / arr[j]</code>.</p>

<p>Return <em>the</em> <code>k<sup>th</sup></code> <em>smallest fraction considered</em>. Return your answer as an array of integers of size <code>2</code>, where <code>answer[0] == arr[i]</code> and <code>answer[1] == arr[j]</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [1,2,3,5], k = 3
<strong>Output:</strong> [2,5]
<strong>Explanation:</strong> The fractions to be considered in sorted order are:
1/5, 1/3, 2/5, 1/2, 3/5, and 2/3.
The third fraction is 2/5.
</pre>

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

<pre>
<strong>Input:</strong> arr = [1,7], k = 1
<strong>Output:</strong> [1,7]
</pre>

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

<ul>
	<li><code>2 &lt;= arr.length &lt;= 1000</code></li>
	<li><code>1 &lt;= arr[i] &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>arr[0] == 1</code></li>
	<li><code>arr[i]</code> is a <strong>prime</strong> number for <code>i &gt; 0</code>.</li>
	<li>All the numbers of <code>arr</code> are <strong>unique</strong> and sorted in <strong>strictly increasing</strong> order.</li>
	<li><code>1 &lt;= k &lt;= arr.length * (arr.length - 1) / 2</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Can you solve the problem with better than <code>O(n<sup>2</sup>)</code> complexity?
class Solution {
public:
    vector<int> kthSmallestPrimeFraction(vector<int>& arr, int k) {

        priority_queue< pair<double,vector<int>> , vector<pair<double,vector<int>>> , greater<pair<double,vector<int>>> > pq;
        set< pair<int,int> > ayegandu; 
        for(int i=1;i<arr.size();i++){
            vector<int> temp = {0,i};
            pq.push(make_pair(1.0/arr[i],temp));
            ayegandu.insert(make_pair(temp[0],temp[1]));
        }

        
        vector<int> ans;
        
        while(k--){
            // cout<<"------------------------------------------------------"<<endl;;
            ans= pq.top().second;
            // cout<<"Popping this element: "<< pq.top().first<<"     "<<pq.top().second[0]<<" "<<pq.top().second[1]<<endl;
            ayegandu.erase({pq.top().second[0],pq.top().second[1]});
            pq.pop();

            double new_num = arr[1+ans[0]];
            double new_den = arr[ans[1]];
            double new_num_1 =arr[ans[0]];
            double new_den_1 = arr[ans[1]-1];

            
            
                vector<int> temp = {1+ans[0],ans[1]};
                // cout<<"Fraction to push: "<<new_num/new_den<<endl;

                if(ayegandu.find({1+ans[0],ans[1]})==ayegandu.end()){
                    // cout<<"Pushing this element: "<<new_num/new_den<<"    "<<temp[0]<<"       "<<temp[1]<<"       "<<endl;
                    pq.push(make_pair(new_num/new_den,temp));
                    ayegandu.insert({1+ans[0],ans[1]});
                }
            
            
                temp = {ans[0],ans[1]-1};
                // cout<<new_num_1/new_den_1<<"    "<<temp[0]<<"       "<<temp[1]<<"       "<<endl;
                // cout<<"Fraction to push: "<<new_num_1/new_den_1<<endl;
                if(ayegandu.find({ans[0],ans[1]-1})==ayegandu.end()){
                    // cout<<"Pushing this element: "<<new_num_1/new_den_1<<"    "<<temp[0]<<"       "<<temp[1]<<"       "<<endl;
                    pq.push(make_pair(new_num_1/new_den_1,temp));
                    ayegandu.insert({ans[0],ans[1]-1});
                
            }
        }
        vector<int> shakala =  ans;
        vector<int> maybe = {arr[shakala[0]],arr[shakala[1]]};

        return maybe;
    }
};
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