<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 maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
        int n = matrix.size();
        int m = matrix[0].size();
        std::vector<std::vector<int>> prefix_sums(n, std::vector<int>(m, 0));
        
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                prefix_sums[i][j] += matrix[i][j];
                if(j > 0) {
                    prefix_sums[i][j] += prefix_sums[i][j-1];
                }
            }
        }

        int best = numeric_limits<int>::min();
        for(int j_start = 0; j_start < m; j_start++) {
            for(int j_end = j_start; j_end < m; j_end++) {
                set<int> tree{};
                int cum_sum = 0;
                for(int i = 0; i < n; i++) {
                    int current = prefix_sums[i][j_end];
                    if(j_start > 0) {
                        current -= prefix_sums[i][j_start-1];
                    }
                    cum_sum += current;

                    auto it = tree.upper_bound(cum_sum - k - 1);
                    if(it != tree.end()) {
                        best = max(best, cum_sum - *it);
                    } 
                    if(cum_sum <= k) {
                        best = max(best, cum_sum);
                    }

                    tree.insert(cum_sum);

                    if(best == k) {
                        return k;
                    }
                }
            }
        }

        return best;
    }
};
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