<p>Given an <code>m x n</code> matrix <code>matrix</code> and an integer <code>k</code>, return <em>the max sum of a rectangle in the matrix such that its sum is no larger than</em> <code>k</code>.</p>

<p>It is <strong>guaranteed</strong> that there will be a rectangle with a sum no larger than <code>k</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/18/sum-grid.jpg" style="width: 255px; height: 176px;" />
<pre>
<strong>Input:</strong> matrix = [[1,0,1],[0,-2,3]], k = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> Because the sum of the blue rectangle [[0, 1], [-2, 3]] is 2, and 2 is the max number no larger than k (k = 2).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[2,2,-1]], k = 3
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 100</code></li>
	<li><code>-100 &lt;= matrix[i][j] &lt;= 100</code></li>
	<li><code>-10<sup>5</sup> &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> What if the number of rows is much larger than the number of columns?</p>

class Solution:
    def maxSumSubmatrix(self, matrix, k):
        rows, cols = len(matrix), len(matrix[0])
        max_sum = float('-inf')
        
        #loop over all pairs of rows
        for top in range(rows):
            #initialize a 1D array to store the sum of elements between the 'top' row and the 'bottom' row
            col_sums = [0] * cols
            
            #expand the window downwards by increasing 'bottom'
            for bottom in range(top, rows):
                #add the values from the 'bottom' row to the col_sums
                for col in range(cols):
                    col_sums[col] += matrix[bottom][col]
                
                #find the maximum subarray sum in col_sums such that the sum is <= k
                curr_sum = 0
                prefix_sums = [0]
                local_max = float('-inf')
                
                for sum_val in col_sums:
                    curr_sum += sum_val
                    
                    #find the smallest prefix_sum such that curr_sum - prefix_sum <= k
                    for prefix_sum in prefix_sums:
                        if curr_sum - prefix_sum <= k:
                            local_max = max(local_max, curr_sum - prefix_sum)
                    
                    prefix_sums.append(curr_sum)
                
                max_sum = max(max_sum, local_max)
        
        return max_sum

17000