<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> ksmallest(vector<int>& a, vector<int>& b, int k) {
        vector<int> ans;
        vector<vector<int>> h;

        for(int i=0; i<a.size(); i++) {
            h.push_back(vector<int>{-(a[i]+b[0]), 0});
        }

        make_heap(h.begin(), h.end());

        while(k>0 && h.size()>0) {
            vector<int> t = h[0];
            ans.push_back(-t[0]);
            pop_heap(h.begin(), h.end());
            h.pop_back();

            if(t[1]+1<b.size()) {
                h.push_back(vector<int>{t[0]+b[t[1]]-b[t[1]+1], t[1]+1});
                push_heap(h.begin(), h.end());
            }
            k--;
        }

        return ans;
    }
    
    int kthSmallest(vector<vector<int>>& mat, int k) {
        vector<int>& v = mat[0];

        for(int i=1; i<mat.size(); i++) {
            v = ksmallest(v, mat[i], k);
        }

        return v[k-1];
    }
};
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