<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 {
public:
  

int findmax(vector<int>& mat, int k) {
    int sum = 0, maxSum = INT_MIN;
    set<int> prefixSums;
    prefixSums.insert(0);  // To handle the case where the subarray starts from the beginning

    for (int val : mat) {
        sum += val;
        
        // Find the smallest prefixSum such that sum - prefixSum <= k
        auto it = prefixSums.lower_bound(sum - k);
        if (it != prefixSums.end()) {
            maxSum = max(maxSum, sum - *it);
        }
        
        // Insert the current sum into the set of prefix sums
        prefixSums.insert(sum);
    }

    return maxSum;
}

int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
    int n = matrix.size();
    int m = matrix[0].size();
    int maxi = INT_MIN;

    // Iterate over all possible starting rows
    for (int top = 0; top < n; ++top) {
        vector<int> mat(m, 0);  // Initialize/reset column sums
        
        // Iterate over all rows from top to bottom
        for (int bottom = top; bottom < n; ++bottom) {
            // Update column sums between the top and bottom rows
            for (int j = 0; j < m; ++j) {
                mat[j] += matrix[bottom][j];
            }
            
            // Find the maximum sum of a submatrix with sum <= k
            maxi = max(maxi, findmax(mat, k));
        }
    }

    return maxi;
}

};
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