<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>

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, std::less<T>, rb_tree_tag, tree_order_statistics_node_update>;
class Solution {
public:
    int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
        int m  = matrix.size();
        int n = matrix[0].size() ;
        vector<vector<int>> pref;
        pref.resize(m+1,vector<int>(n+1,0));

        for(int i=0;i<m;i++){
            pref[i][0]=matrix[i][0];
            for(int j=1;j<n;j++){
                pref[i][j]=pref[i][j-1]+matrix[i][j];
            }
        }
        int ans=-1e9;
        for(int i=0;i<n;i++){
            for(int j=i;j<n;j++){
                int cum=0;
                ordered_set<int> st;
                st.insert(0);


                for(int l=0;l<m;l++){
                    int sum_row= pref[l][j];
                    if(i!=0){
                        sum_row-=pref[l][i-1];
                    }


                    cum+=sum_row;
                    auto it = st.order_of_key(cum - k);
                    if (it != st.size()) {
                        ans = max(ans, cum - *st.find_by_order(it));
                    }
                    st.insert(cum);



                }
            }
        }

        return ans;
    }
};
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