<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>>& a, int k) {

        int n = a.size();
        int m = a[0].size();

        int ans = INT_MIN;

        vector<vector<int>> sums(n, vector<int> (m, 0));
        int sum = 0;
        int rowSum = 0;

        for (int j = 0; j < m; j++) {
            sum = 0;
            for (int i = 0; i < n; i++) {
                sum += a[i][j];
                sums[i][j] = sum;
            }
        }

        for (int l = 0; l < n; l++) {
            for (int r = l; r < n; r++) {
                sum = 0;
                set<int> s;
                s.insert(0);
                for (int j = 0; j < m; j++) {
                    rowSum = sums[r][j];
                    if(l > 0) rowSum -= sums[l - 1][j];
                    sum += rowSum;
                    auto lb = s.lower_bound(sum - k);
                    if(lb != s.end()) ans = max(ans, sum - *lb);
                    s.insert(sum);
                }
            }
        }

        return ans;

    }
};
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