<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>

import heapq

class Solution:
    def kthSmallest(self, mat: List[List[int]], k: int) -> int:
        n_row = len(mat)
        n_col = len(mat[0])
        
        if n_col == 1:
            return sum((mat[i][0] for i in range(n_row)))
        
        h = []
        # h is a heap
        # each entry is (total sum, list of column indices)
        
        start_indices = [0] * n_row
        start_sum = sum((mat[i][0] for i in range(n_row)))
        heappush(h, (start_sum, start_indices))
        
        visited = set()
        visited.add(str(start_indices))
        
        for ki in range(k):
            list_sum, indices = heappop(h)
            if ki == k - 1:
                return list_sum
            
            for i in range(n_row):
                j = indices[i]
                if j < n_col - 1:
                    new_list = indices.copy()
                    new_list[i] = j + 1
                    
                    key = str(new_list)
                    if key in visited:
                        continue
                    visited.add(key)
                    
                    new_sum = list_sum - mat[i][j] + mat[i][j+1]
                    heappush(h, (new_sum, new_list))
            
        return -1
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