<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:
    def kthSmallest(self, mat: List[List[int]], k: int) -> int:
        m, n = len(mat), len(mat[0])
        
        # Start with the smallest sum using the first element of each row
        min_heap = [(sum(row[0] for row in mat), [0] * m)]  # (sum, indexes in each row)
        seen = set()
        seen.add(tuple([0] * m))
        
        # Perform k iterations to find the k-th smallest sum
        for _ in range(k):
            curr_sum, indices = heapq.heappop(min_heap)
            
            # If it's the k-th iteration, return the sum
            if _ == k - 1:
                return curr_sum
            
            # Explore the next possible sums by moving one element to the right in any row
            for i in range(m):
                new_indices = indices[:]
                if new_indices[i] + 1 < n:
                    new_indices[i] += 1
                    new_tuple = tuple(new_indices)
                    if new_tuple not in seen:
                        seen.add(new_tuple)
                        new_sum = curr_sum - mat[i][indices[i]] + mat[i][new_indices[i]]
                        heapq.heappush(min_heap, (new_sum, new_indices))

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