<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>>& m, int k) {
        int res=INT_MIN,rows=m.size(),cols=m[0].size();
        for(int l=0;l<cols;++l){
            vector<int>sums(rows);
            for(int r=l;r<cols;++r){
                int kadane=0,max_kadane=INT_MIN;
                for(int i=0;i<rows;++i){
                    sums[i]+=m[i][r];
                    kadane=max(kadane+sums[i],sums[i]);
                    max_kadane=max(max_kadane,kadane);
                }
                if(max_kadane<=k){
                    res=max(res,max_kadane);
                    continue;
                }
                set<int> s = {0};
            int run_sum = 0;
            for (int sum : sums) {
                run_sum += sum;
                auto it = s.lower_bound(run_sum - k);
                if (it != end(s))
                    res = max(res, run_sum - *it);
                s.insert(run_sum);
            }
            }
        }
        return res;
    }
};
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