<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 m =matrix.size();
        int n = matrix[0].size();
        for(int i=1;i<m;++i){
            for(int j=0;j<n;++j){
                matrix[i][j] += matrix[i-1][j];
                // cout<<matrix[i][j]<<" ";
            }
        }
        
        int ans = INT_MIN ;
        
        
        
        for(int i=0;i<m;++i){
            for(int j=i;j<m;++j){
                vector<int> temp = matrix[j];
                if(i!=0){
                    for(int t=0;t<n;++t){
                        temp[t]-=matrix[i-1][t];
                    }
                }
                // cout<<" Hloe"<<endl;
                
                //find greatest sum <= k in a 1d array O(nlog(n))

                set<int> st;
                int prefix =0 ;
                // st.insert(0);
                
                for(int t=0;t<n;++t){
                    st.insert(prefix);
                    prefix+=temp[t];
                    auto it = lower_bound(st.begin(), st.end(), (prefix - k) );
                    if(it!= st.end()){
                        if(prefix - *it >= ans ){
                            ans = prefix - *it;
                        }
                    }
                    // st.insert(prefix);
                }
            }
        }
        
        
        
        return ans;
    }
};
226320