<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:
    string stringify(vector <int> curr){
        return to_string(curr[0]) + "," + to_string(curr[1]);
    }
    vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        priority_queue <pair <int, vector <int>>, vector<pair <int, vector <int>>>, greater<pair <int, vector <int>>>> q;
        vector <vector <int>> ans; int taken = 0;
        q.push({nums1[0] + nums2[0], {0, 0}});
        unordered_set <string> visited;
        while (!q.empty() && taken < k){
            vector <int> curr_ind = q.top().second; 
            q.pop();
            //cout << stringify(curr_ind) << endl;
            if (visited.find(stringify(curr_ind)) != visited.end()){
                continue;
            }
            ans.push_back({nums1[curr_ind[0]], nums2[curr_ind[1]]}); 
            visited.insert(stringify(curr_ind));
            if (curr_ind[0] < nums1.size() - 1){
                q.push({nums1[curr_ind[0] + 1] + nums2[curr_ind[1]], {curr_ind[0] + 1, curr_ind[1]}});
            }
            if (curr_ind[1] < nums2.size() - 1){
                q.push({nums1[curr_ind[0]] + nums2[curr_ind[1] + 1], {curr_ind[0], curr_ind[1] + 1}});
            }
            taken++;
        }
        return ans;
    }
};


//1,2
//1,4 / 7,2
//7,2 / 1,6 / 7,4  --> 1,6 / 7,2 / 7,4
//7,6 / 7,2 / 7,4

//1,1
//1,2 / 1,1 --> 1,1 / 1,2
//
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