<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>> const& matrix, int k) {
        int m = matrix.size(), n = matrix[0].size(), ans{INT_MIN};
        vector<int> sum(n);
        for (int l{}; l < m; ++l) {
            fill(sum.begin(), sum.end(), 0);
            for (int r{l}; r < m; ++r) {
                int cur{}, maxsum{INT_MIN};
                for (int c{}; c < n; ++c) {
                    sum[c] += matrix[r][c];
                    cur = max(sum[c], cur + sum[c]);
                    maxsum = max(cur, maxsum);
                }
                if (maxsum <= k) {
                    ans = max(ans, maxsum);
                    continue;
                }
                set<int> s;
                s.insert(0);
                cur = 0;
                for (int c{}; c < n; ++c) {
                    cur += sum[c];
                    if (auto it{lower_bound(s.begin(), s.end(), cur - k)}; it != s.end()) {
                        ans = max(ans, cur - *it);
                    }
                    s.insert(cur);
                }
            }
        }
        return ans;
    }
};
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