<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) {
        vector<pair<double, pair<int,int>>> p;
        for(int i = 0; i <arr.size()-1; i++){
            for(int j = i+1; j < arr.size(); j++){
                pair<int,int> ps;
                ps.first = arr[i];
                ps.second = arr[j];
                pair<double, pair<int, int>> pd;
                pd.first = (double)arr[i]/arr[j];
                pd.second = ps;
                p.push_back(pd);
            }
        }

 auto cmp = [](const pair<double, pair<int, int>>& a, 
                  const pair<double, pair<int, int>>& b) {
        return a.first > b.first; 
    };

    priority_queue<pair<double, pair<int, int>>, vector<pair<double, pair<int, int>>>, decltype(cmp)> pq(p.begin(), p.end(), cmp);
    k--;
    while(k--){
        pq.pop();
    }
    auto[one, two] = pq.top().second;
    return {one, two};

    }
};
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