<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>

struct T {
  int i;
  int j;
  int sum;  // nums1[i] + nums2[j]
};

class Solution {
 public:
  int kthSmallest(vector<vector<int>>& mat, int k) {
    vector<int> row = mat[0];

    for (int i = 1; i < mat.size(); ++i)
      row = kSmallestPairSums(row, mat[i], k);

    return row.back();
  }

 private:
  // Similar to 373. Find K Pairs with Smallest Sums
  vector<int> kSmallestPairSums(vector<int>& nums1, vector<int>& nums2, int k) {
    vector<int> ans;
    auto compare = [&](const T& a, const T& b) { return a.sum > b.sum; };
    priority_queue<T, vector<T>, decltype(compare)> minHeap(compare);

    for (int i = 0; i < k && i < nums1.size(); ++i)
      minHeap.emplace(i, 0, nums1[i] + nums2[0]);

    while (!minHeap.empty() && ans.size() < k) {
      const auto [i, j, _] = minHeap.top();
      minHeap.pop();
      ans.push_back(nums1[i] + nums2[j]);
      if (j + 1 < nums2.size())
        minHeap.emplace(i, j + 1, nums1[i] + nums2[j + 1]);
    }

    return ans;
  }
};
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