<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:
        nRows, nCols = len(mat), len(mat[0])
        minSum, maxSum = 0, 0
        for r in range(nRows):
            minSum += mat[r][0]
            maxSum += mat[r][nCols-1]
        
        def count(row,curSum,val):
            if curSum > val:
                return 0
            if row == nRows:
                return 1
            ans = 0
            for col in range(nCols):
                if curSum + mat[row][col] > val:
                    break
                countNxt = count(row+1,curSum+mat[row][col],val)
                if countNxt == 0:
                    break
                ans += countNxt
                if ans > k:
                    break
            return ans

        
        l,r = minSum, maxSum
        while l <= r:
            m = l + (r-l)//2
            ct = count(0,0,m)
            if ct < k:
                l = m+1
            else:
                r = m-1
        return l
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