<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:
        from heapq import heapify
        sumi = [[0,[]]] # sum total, [indices of locations]
        m, n = len(mat), len(mat[0])
        for row in mat:
            sumi[0][0]+=row[0]
            sumi[0][1].append(0)
        candidates = {} # it has (tuple of indices): result sum of those vals
        visited = set([tuple(sumi[0][1])])
        while len(sumi)<k:
            #print(sumi, visited)
            min_sum = 1e10
            best_indices = []
            found = False
            for val, cur_indices in sumi:
                if found ==True:
                    break
                for row_idx in range(m):# the row to step forward
                    new_indices = [cur_indices[i] if (i != row_idx or cur_indices[i]==n-1) else cur_indices[i]+1 for i in range(m) ]      
                    
                    if cur_indices[row_idx]==n-1:
                        continue
                    new_val = val + mat[row_idx][cur_indices[row_idx]+1]-mat[row_idx][cur_indices[row_idx]]
                    if tuple(new_indices) in visited:
                        #print('skipping', new_indices)
                        continue
                    if new_val == val:
                        found = True
                        min_sum = new_val
                        best_indices = new_indices
                        break
                    if new_val<min_sum:
                        min_sum = new_val
                        best_indices = new_indices
            # we found our best next step
            sumi.append([min_sum, best_indices])
            visited.add(tuple(best_indices))
        #print(sumi)
        return sumi[-1][0]
17600