<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?
#pragma GCC target("avx, mmx, sse2, sse3, sse4")

auto disableSync = [](void) noexcept -> int
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    return 0;
}();

class Solution final
{
public:
    std::vector<int> kthSmallestPrimeFraction(const std::vector<int>& primes, const int k) const noexcept
    {
        const size_t nbPrimes = primes.size();
        std::vector<std::pair<int, int>> fractions;
        fractions.reserve(nbPrimes * (nbPrimes - 1) / 2 + 10);
        for (auto it1 = primes.cbegin(); it1 != primes.cend(); ++it1)
            for (auto it2 = it1 + 1; it2 != primes.cend(); ++it2)
                fractions.push_back(std::make_pair(*it1, *it2));
        std::nth_element(fractions.begin(), fractions.begin() + k - 1, fractions.end(),
            [](const std::pair<int, int>& a, const std::pair<int, int>& b) noexcept -> bool
                {
                    return (double)a.first / a.second < (double)b.first / b.second;
                });
        return {fractions[k - 1].first, fractions[k - 1].second};
    }
};
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