<p>You are given two integer arrays <code>nums1</code> and <code>nums2</code> sorted in <strong>non-decreasing&nbsp;order</strong> and an integer <code>k</code>.</p>

<p>Define a pair <code>(u, v)</code> which consists of one element from the first array and one element from the second array.</p>

<p>Return <em>the</em> <code>k</code> <em>pairs</em> <code>(u<sub>1</sub>, v<sub>1</sub>), (u<sub>2</sub>, v<sub>2</sub>), ..., (u<sub>k</sub>, v<sub>k</sub>)</code> <em>with the smallest sums</em>.</p>

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

<pre>
<strong>Input:</strong> nums1 = [1,7,11], nums2 = [2,4,6], k = 3
<strong>Output:</strong> [[1,2],[1,4],[1,6]]
<strong>Explanation:</strong> The first 3 pairs are returned from the sequence: [1,2],[1,4],[1,6],[7,2],[7,4],[11,2],[7,6],[11,4],[11,6]
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [1,1,2], nums2 = [1,2,3], k = 2
<strong>Output:</strong> [[1,1],[1,1]]
<strong>Explanation:</strong> The first 2 pairs are returned from the sequence: [1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3]
</pre>

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

<ul>
	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums1[i], nums2[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>nums1</code> and <code>nums2</code> both are sorted in <strong>non-decreasing order</strong>.</li>
	<li><code>1 &lt;= k &lt;= 10<sup>4</sup></code></li>
	<li><code>k &lt;=&nbsp;nums1.length *&nbsp;nums2.length</code></li>
</ul>

class Solution {
public:
    vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        int listLength =  nums1.size();

        priority_queue<pair<int, pair<int, int>>,
                    vector<pair<int, pair<int, int>>>, 
                    greater<pair<int, pair<int, int>>>>minHeap;

        vector<vector<int>> pairs;

        for(int i = 0; i < min(k, listLength); i++) {
          minHeap.push({nums1[i] + nums2[0], {i, 0}});
        }

        int counter = 0;

        while(!minHeap.empty() && counter < k) {
          int sumOfPairs = minHeap.top().first;
          int i = minHeap.top().second.first;
          int j = minHeap.top().second.second;
          minHeap.pop();

          pairs.push_back({nums1[i], nums2[j]});
          int nextElement = j + 1;

            // if next element is available for nums2 then add it to the heap
            if (nums2.size() > nextElement) {
                minHeap.push({nums1[i] + nums2[nextElement], {i, nextElement}});
            }

            counter++;
        }

        return pairs;
    }
};
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