<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>

auto init = []() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    return 'c';
}();

class Solution {
public:
    vector<vector<int>> kSmallestPairs(vector<int>& n1, vector<int>& n2, int k) {
        int n=n1.size();
        int m=n2.size();
        int i=0; int j=0;
        priority_queue<vector<int>> pq;
        vector<int> v(2);
        while(k--){
            v[0]=n1[i]+n2[j];
            v[1]=i*m+j;
            pq.push(v);
            j++;
            if(j==m){
                i++;
                j=0;
            }
        }
        for(i; i<n; i++){
            while(true){
                if(j>=m){
                    j=0;
                    break;
                }
                if(n1[i]+n2[j]<pq.top()[0]){
                    v[0]=n1[i]+n2[j];
                    v[1]=i*m+j;
                    pq.push(v);
                    pq.pop();
                }
                else{
                    j=0;
                    break;
                }
                j++;
            }
        }
        vector<vector<int>> ans;
        while(!pq.empty()){
            v[0]=n1[pq.top()[1]/m];
            v[1]=n2[pq.top()[1]%m];
            ans.push_back(v);
            pq.pop();
        }
        return ans;
    }
};
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