<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) {
        // kth smallest in m * n matrix comes from merging kth smallest in (m - 1) * n matrix with the last row and trimming to k elements
        // dp[i] = {k smallest sums in using 1st i rows}
        // dp[0] = mat[0][k elements]
        // dp[i] = dp[i - 1] X mat[i] taking k smallest elements (X is cartesian product sum or cross product)
        // ans = dp.back()
        int n = mat.size();
        int m = mat[0].size();
        vector<int> dp;
        for(int i = 0; i < min(m, k); i++) {
            dp.push_back(mat[0][i]);
        }
        for(int i = 1; i < n; i++) {
            vector<int> new_dp;
            multiset<int> ms;
            for(int j = 0; j < m; j++) {
                for(auto x: dp) {
                    if(ms.size() >= k) {
                        if(*ms.rbegin() > x + mat[i][j]) {
                            ms.erase(ms.find(*ms.rbegin()));
                            ms.insert(x + mat[i][j]);
                        }
                    }
                    else {
                        ms.insert(x + mat[i][j]);
                    }
                }
            }
            for(auto x: ms) {
                new_dp.push_back(x);
            }
            swap(dp, new_dp);
        }
        return dp.back();
    }
};
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