<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 subArraySumEqualsK(vector<int> &sum,int k){
        int n=sum.size();
        set<int> s;
        
        s.insert(0);
        
        int res= INT_MIN;
        int prefixSum=0;
        for(int i=0;i<n;i++){
            prefixSum+=sum[i];
            
            auto it= s.lower_bound(prefixSum-k);
            if(it != s.end()){                     // ( (prefixSum) - ((prefixSum-k)+lambda) )
                res=max(res,prefixSum - *it);      //  k-lambda
            }
            if(res==k)return k;
            
            s.insert(prefixSum);
        }
        
        return res;
    }
    int maxSumSubArray(vector<int> &sum){
        int n=sum.size();
        
        int prefixSum=0;
        int maxSum= INT_MIN;
        
        for(int i=0;i<n;i++){
            
            prefixSum+=sum[i];
            maxSum=max(maxSum,prefixSum);
            
            if(prefixSum<0)prefixSum=0;
        }
        
        return maxSum;
    }
    int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
        int n= matrix.size();
        int m= matrix[0].size();
        
        int res=INT_MIN;
        
        for(int j=0;j<m;j++){
            
                vector<int> sum(n,0);

            for(int startCol=j;startCol<m;startCol++){
                
                for(int i=0;i<n;i++){
                    
                    sum[i]+= matrix[i][startCol];
                }
                
             int currMax=maxSumSubArray(sum);
                
            if(currMax==k)return k;
            
            else if(currMax<k){
                res=max(res,currMax);
            }
            
            else{
                
                currMax= subArraySumEqualsK(sum,k);
                res=max(res,currMax);
            }
                
            }
        }
        
        return res;
        
    }
};
35128