<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> kSmallestSumPair(vector<int>& a, vector<int>& b, int k) {
    vector<int> ans;
    priority_queue<int> pq;
    for (int i = 0; i < a.size(); i++) {
      for (int j = 0; j < b.size(); j++) {
        pq.push(a[i] + b[j]);
        if (pq.size() > k) {
          pq.pop();
        }
      }
    }
    while (!pq.empty()) {
      ans.push_back(pq.top());
      pq.pop();
    }
    
    return ans;
  }
  
  int kthSmallest(vector<vector<int>>& mat, int k) {
    //sort(mat.begin(), mat.end());
    
    int m = mat.size();
    int n = mat.front().size();
    
    if (m == 1) return mat[0][k - 1];
    
    vector<int> ans = kSmallestSumPair(mat[0], mat[1], k);
    
    for (int i = 2; i < m; i++) {
      ans = kSmallestSumPair(ans, mat[i], k);
    }
    
    return ans[0];    
  }
};
15383