<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>

class Solution {
public:
    vector<int>f(vector<int>mat,vector<int>res, int k) {
        // finding the k smallest pairs for the combination of theses 2 arrays
        priority_queue<vector<int>,vector<vector<int>>,greater<vector<int>>>pq;
        int n =mat.size();
        for (int i=0;i<res.size();i++) {
            pq.push({res[i]+mat[0],0,i});
        }
        vector<int>ans;
        while (!pq.empty() && k--) {
            vector<int>cur = pq.top();
            pq.pop();
            ans.push_back(mat[cur[1]]+res[cur[2]]);
            if (cur[1]+1<n){
                pq.push({mat[cur[1]+1]+res[cur[2]],cur[1]+1,cur[2]});
            }
        }
        return ans;
    }
    int kthSmallest(vector<vector<int>>& mat, int k) {
        int n = mat.size();
        if (n==1) return mat[0][k-1];
        vector<int>res = f(mat[0],mat[1],k);
        for (int i=2;i<n;i++) {
            res = f(res,mat[i],k);
        }
        return res[k-1];
    }
};
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