<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:
    int kthSmallest(vector<vector<int>>& mat, int k) {
        vector<int>& total = mat[0];
        for (int i=1; i<mat.size(); i++) {
            total = kthSmallestPairs(total, mat[i], k);
        }
        return total[k-1];
    }

    vector<int> kthSmallestPairs(vector<int>& A, vector<int>& B, int k) {
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
        set<pair<int, int>> visited;
        pq.push({A[0]+B[0], 0, 0});
        visited.insert({0,0});
        vector<int> res;

        while (!pq.empty() && k > 0) {
            int pairTotal = pq.top()[0];
            int i = pq.top()[1];
            int j = pq.top()[2];
            pq.pop();
            --k;
            res.push_back(pairTotal);

            if (i+1<A.size() && visited.find({i+1,j}) == visited.end()) {
                pq.push({A[i+1]+B[j], i+1, j});
                visited.insert({i+1, j});
            }
            if (j+1<B.size() && visited.find({i,j+1}) == visited.end()) {
                pq.push({A[i]+B[j+1], i, j+1});
                visited.insert({i, j+1});
             }
        }

        return res;
    }
};
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