<p>You are given an <code>m x n</code> matrix <code>mat</code> that has its rows sorted in non-decreasing order and an integer <code>k</code>.</p>

<p>You are allowed to choose <strong>exactly one element</strong> from each row to form an array.</p>

<p>Return <em>the </em><code>k<sup>th</sup></code><em> smallest array sum among all possible arrays</em>.</p>

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

<pre>
<strong>Input:</strong> mat = [[1,3,11],[2,4,6]], k = 5
<strong>Output:</strong> 7
<strong>Explanation:</strong> Choosing one element from each row, the first k smallest sum are:
[1,2], [1,4], [3,2], [3,4], [1,6]. Where the 5th sum is 7.
</pre>

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

<pre>
<strong>Input:</strong> mat = [[1,3,11],[2,4,6]], k = 9
<strong>Output:</strong> 17
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> mat = [[1,10,10],[1,4,5],[2,3,6]], k = 7
<strong>Output:</strong> 9
<strong>Explanation:</strong> Choosing one element from each row, the first k smallest sum are:
[1,1,2], [1,1,3], [1,4,2], [1,4,3], [1,1,6], [1,5,2], [1,5,3]. Where the 7th sum is 9.  
</pre>

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

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat.length[i]</code></li>
	<li><code>1 &lt;= m, n &lt;= 40</code></li>
	<li><code>1 &lt;= mat[i][j] &lt;= 5000</code></li>
	<li><code>1 &lt;= k &lt;= min(200, n<sup>m</sup>)</code></li>
	<li><code>mat[i]</code> is a non-decreasing array.</li>
</ul>

/*
[[1,10,10],
[1,4,5],
[2,3,6]]
*/

class Solution {
public:
    
    vector<int> kSmallestPairs(vector<int>& a, vector<int>& b, int k = 200){
        priority_queue<pair<int, pair<int, int>>> pq;
        int n = a.size();
        int m = b.size();
        for(int i = 0; i < b.size() && n; i++){
            pq.push({-1 * (a[0] + b[i]) , {0, i}});
        }
        vector<int> res;
        map<pair<int, int>, int> vis;
        for(int i = 0; i < k && n && m; i++){
            int x = pq.top().second.first;
            int y = pq.top().second.second;
            if(!vis.count({x, y})){
                res.push_back(a[x] + b[y]);
                vis[{x, y}]++;
            }
            pq.pop();
            if(x + 1 < a.size()){
                pq.push({-1 * (a[x + 1] + b[y]), {x + 1, y}});
            }
        }
        return res;
    }
    
    
    int kthSmallest(vector<vector<int>>& a, int k) {
        int n = a.size();
        vector<int> res = a[0];
        for(int i = 1; i < n; i++){
            res = kSmallestPairs(res, a[i]);
        }
        return res[k - 1];
    }
};
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