<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>>& grid, int target) {
        int n = grid.size(), m = grid[0].size();
        int res = INT_MIN;
        for (int j=0;j<m;j++) {
            vector<int> arr(n, 0);
            for (int k=j;k>=0;k--) {
                for (int i=0;i<n;i++) {
                    arr[i] += grid[i][k];
                }

                int cursum=0;
                set<int> set;
                set.insert(0);
                for (int i=0;i<n;i++) {
                    cursum += arr[i];
                    auto it = set.lower_bound(cursum - target);
                    if (it != set.end()) {
                        if ((cursum - (*it)) == target) return target;
                        res = max(res, cursum - (*it));
                    }
                    set.insert(cursum);
                }
            }
        }

        return res;
    }
};
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