<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

def hash_indexes(indexes):
    s = ""
    for i in indexes:
        s += str(i)
    return s

class Solution:
    def kthSmallest(self, mat: List[List[int]], k: int) -> int:
        n, m = len(mat), len(mat[0])
        pq = []
        curr = 0
        for i in range(n):
            curr += mat[i][0]
        pq.append((curr, [0]*n))
        heapq.heapify(pq)
        ans = -1

        vis = set()
        vis.add('0'*n)
        
        while pq and k:
            curr_sum, indexes = heapq.heappop(pq)
            
            # print(f"curr_sum: {curr_sum}, indexes: {indexes}")
            k -= 1
            if not k:
                ans = curr_sum
                break
            
            for i, ind in enumerate(indexes):
                if ind == m-1:
                    continue
                
                temp = indexes.copy()
                temp[i] += 1 
                # print(temp)
                temp_hash = hash_indexes(temp)
                # print(temp_hash)
                if temp_hash not in vis:
                    heapq.heappush(pq, (curr_sum + mat[i][ind+1] - mat[i][ind], temp))
                    vis.add(temp_hash)

        return ans
        
197