<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>>& matrix, int k) {
        int n=matrix.size();
         int m=matrix[0].size();
        vector<vector<int>> sum(n+1,vector<int>(m+1,0));
        for(int i=1;i<=n;i++){
             for(int j=1;j<=m;j++){
                 sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+matrix[i-1][j-1];
             }
        }
         int ans=-1e5;
        for(int col1=0;col1<m;col1++){
            for(int col2=col1+1;col2<=m;col2++){
               map<int,int> mp ;
               for(int row=1;row<=n;row++){
                int  s1=sum[row][col2]-sum[row][col1];

                mp[0]=0;
                //  if(s1==k)
                //   return k;
                  auto it=mp.lower_bound(s1-k);
                  if(it!=mp.end() && s1-it->first<=k)
                  ans=max(ans,s1-it->first);
                    //  cout<<row<<" "<<s1<<" "<<it->first<<endl ;
                  mp[s1]=row;
               }
            }
        }
        // if(ans>k)
        // return -1 ;
        return ans ;
      

    }
};
180064