<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
from typing import List

class Solution:
    def kthSmallest(self, mat: List[List[int]], k: int) -> int:
        # Number of rows in the matrix
        rows = len(mat)
        
        # Initial min-heap and the initial sum
        heap = []
        initial_sum = sum(row[0] for row in mat)
        initial_indices = tuple(0 for _ in range(rows))
        heapq.heappush(heap, (initial_sum, initial_indices))
        
        # Set to track visited states
        visited = set()
        visited.add(initial_indices)
        
        # We need to extract the k-th smallest element from the heap
        for _ in range(k):
            current_sum, indices = heapq.heappop(heap)
            
            # If it's the k-th smallest sum, return it
            if _ == k - 1:
                return current_sum
            
            # Generate new sums by incrementing the index of one row
            for r in range(rows):
                if indices[r] + 1 < len(mat[r]):
                    # Generate new indices tuple
                    new_indices = list(indices)
                    new_indices[r] += 1
                    new_indices = tuple(new_indices)
                    
                    # Calculate the new sum
                    new_sum = current_sum - mat[r][indices[r]] + mat[r][new_indices[r]]
                    
                    if new_indices not in visited:
                        visited.add(new_indices)
                        heapq.heappush(heap, (new_sum, new_indices))
        
        # In case something goes wrong (should not happen if k is valid)
        return -1

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