<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 kadane(vector<int>&arr,int k){
        int curr_sum = 0;
        int max_sum = INT_MIN;
        set<int>s;
        s.insert(0);
        for(int i=0;i<arr.size();i++){
            curr_sum+=arr[i];
            auto it = s.lower_bound(curr_sum-k);
            if(it!=s.end()){
                max_sum = max(curr_sum - *it,max_sum);
            }
            s.insert(curr_sum);
        }
        return max_sum;
    }
    int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
        int r=matrix.size();
        int c=matrix[0].size();
        int ans=INT_MIN;
        for(int col_start = 0; col_start < c; col_start++){
            vector<int>temp(r,0);
            for(int col_end = col_start;col_end<c;col_end++){
                for(int row=0;row<r;row++){
                    temp[row]+=matrix[row][col_end];
                }
                int curr_max_sum = kadane(temp,k);
                ans = max(ans,curr_max_sum);
            }
        }
        return ans;
    }
};
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