<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>

#include <iostream>
#include <vector>
#include <set>
#include <climits>
using namespace std;

class Solution {
public:
    int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
        int rows = matrix.size();
        int cols = matrix[0].size();
        int result = INT_MIN;

        // Iterate over pairs of rows
        for (int top = 0; top < rows; ++top) {
            vector<int> colSums(cols, 0);  // This will store sum of elements in columns between row 'top' and 'bottom'
            
            for (int bottom = top; bottom < rows; ++bottom) {
                // Update the column sums for the current row pair (top, bottom)
                for (int col = 0; col < cols; ++col) {
                    colSums[col] += matrix[bottom][col];
                }
                
                // Now find the maximum subarray sum no larger than k for colSums
                set<int> prefixSums;
                prefixSums.insert(0);  // For subarrays starting from the first element
                int currSum = 0;

                for (int sum : colSums) {
                    currSum += sum;

                    // We need to find the smallest prefixSum such that currSum - prefixSum <= k
                    auto it = prefixSums.lower_bound(currSum - k);

                    if (it != prefixSums.end()) {
                        result = max(result, currSum - *it);
                    }

                    // Insert the current sum into the prefix sums set
                    prefixSums.insert(currSum);
                }
            }
        }

        return result;
    }
};

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