<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?
#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>

using namespace std;

// Custom hash function for pair<int, int>
struct hash_pair {
    template <class T1, class T2>
    size_t operator()(const pair<T1, T2>& p) const {
        auto hash1 = hash<T1>{}(p.first);
        auto hash2 = hash<T2>{}(p.second);
        return hash1 ^ hash2;
    }
};

class Solution {
public:
    pair<int, int> findPairByValue(const unordered_map<pair<int, int>, double, hash_pair>& mpp, double value) {
        for (const auto& entry : mpp) {
            if (entry.second == value) {
                return entry.first;
            }
        }
        return {-1, -1}; // Special value indicating not found
    }

    vector<int> kthSmallestPrimeFraction(vector<int>& arr, int k) {
        unordered_map<pair<int, int>, double, hash_pair> mpp;
        vector<double> frac;

        // Calculate all possible fractions and store in map and vector
        for (int i = 0; i < arr.size() - 1; ++i) {
            for (int j = i + 1; j < arr.size(); ++j) { // Ensure n1 < n2
                int n1 = arr[i];
                int n2 = arr[j];
                double f = static_cast<double>(n1) / n2; // Correctly calculate the fraction
                mpp[{n1, n2}] = f;
                frac.push_back(f);
            }
        }

        // Sort the fractions
        sort(frac.begin(), frac.end());

        // Find the k-th smallest fraction
        double ans = frac[k - 1];
        pair<int, int> p = findPairByValue(mpp, ans);

        // Return the result as a vector
        return {p.first, p.second};
    }
};


296799