<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 helper(int row1,int row2,int m,int k,vector<vector<int>>&dp){
    int ans=INT_MIN;
    int sum=0;
    set<int>st;
    st.insert(0);
    for(int i=0;i<m;i++){
        int val=dp[row2][i]-dp[row1-1][i];
        sum+=val;
        // cout<<sum<<endl;
        int x=sum-k;
        //  cout<<x<<endl;
        auto it=st.lower_bound(x);
        // cout<<(*it)<<endl;
        if(it!=st.end()){
            ans=max(ans,(sum-(*it)));
        } 
        st.insert(sum);
      
    }
    return ans;
}
    int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {

        int n=matrix.size();
        int m=matrix[0].size();
        vector<vector<int>>dp(n+1,vector<int>(m,0));

        for(int row=1;row<=n;row++){
            for(int col=0;col<m;col++){
                dp[row][col]=dp[row-1][col]+matrix[row-1][col];
            }
        }
        int ans=INT_MIN;
        for(int row1=1;row1<=n;row1++){
            for(int row2=row1;row2<=n;row2++){
                int val=helper(row1,row2,m,k,dp);
                // cout<<val<<endl;
                if(val<=k) ans=max(ans,val);

            }
        }

        return ans;
        
    }
};
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