<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();

        vector<int> sums(n, 0);
        int ans=-1e9;

        for (int l=0; l<m; l++) {
            for (int i=0; i<n; i++) {
                sums[i]=0;
            }
            for (int r=l; r<m; r++) {
                for (int i=0; i<n; i++) {
                    sums[i] += matrix[i][r];
                }
                set<int>cset;
                cset.insert(0);
                int curSum=0, curMax=-1e9;
                for (int sum:sums) {
                    curSum += sum;
                    auto it = cset.lower_bound(curSum-k);
                    if (it!=cset.end()) {
                        curMax = max(curMax, curSum-*it);
                    }
                    cset.insert(curSum);
                }
                ans = max(curMax, ans);
            }
        }
        return ans;
    }
};
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