K-th Smallest Prime Fraction
<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?
Array; Two Pointers; Binary Search; Sorting; Heap (Priority Queue)
class Solution {
public:
    vector<int> kthSmallestPrimeFraction(vector<int>& arr, int k) {
        using pii = pair<int, int>;
        int n = arr.size();
        auto cmp = [&](const pii& a, const pii& b) {
            return arr[a.first] * arr[b.second] > arr[b.first] * arr[a.second];
        };
        priority_queue<pii, vector<pii>, decltype(cmp)> pq(cmp);
        for (int i = 1; i < n; ++i) {
            pq.push({0, i});
        }
        for (int i = 1; i < k; ++i) {
            pii f = pq.top();
            pq.pop();
            if (f.first + 1 < f.second) {
                pq.push({f.first + 1, f.second});
            }
        }
        return {arr[pq.top().first], arr[pq.top().second]};
    }
};