<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 = len(mat)
        n = len(mat[0])

        pq = []
        visited = set()

        # min sum is to pick the first one from each row
        min_sum = sum([row[0] for row in mat]) 
        # idx are all 0 for each row
        cols = [0] * m

        heappush(pq, (min_sum, cols))
        visited.add(tuple(cols))

        while k > 0:
            curr_sum, cols = heappop(pq)

            k -= 1
            if k == 0:
                return curr_sum 

            # get next combination from last col idx:
            # move 1 col in only row 0 and get col sum
            # move 1 col in only row 1 and get col sum
            # ...

            for i in range(m):
                nxt_cols, nxt_sum = [], 0
                skip = False
                for r, c in enumerate(cols):
                    if r == i: 
                        c += 1
                    
                    if c >= n:
                        # this col combination is invalid
                        skip = True
                    else:
                        nxt_cols.append(c)
                        nxt_sum += mat[r][c]

                if not skip and tuple(nxt_cols) not in visited:
                    visited.add(tuple(nxt_cols))
                    heappush(pq, (nxt_sum, nxt_cols))




        
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