<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>

from bisect import bisect_left, insort

class Solution:
    def maxSumSubmatrix(self, matrix: List[List[int]], k: int) -> int:

        m, n = len(matrix), len(matrix[0])
        maxSum = float('-inf')

        for left in range(n):
            curr = [0] * m  # every row betwen left and right

            for right in range(left, n):
                # populate the row sums w this row, col value
                for i in range(m):
                    curr[i] += matrix[i][right] 
                
                currSum = 0
                sums = [currSum]
                for c in curr:
                    currSum += c

                    i = bisect_left(sums, currSum - k)
                    # if there is a prefix sum <= the ideal prefix sum
                    if i != len(sums):
                        maxSum = max(maxSum, currSum - sums[i])
                    
                    insort(sums, currSum)
                
        return maxSum


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