<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 target) {
        int n = matrix.size(), m = matrix[0].size();
        int ans = INT_MIN;
        for(int i = 0; i < n;i++){
            for(int j = 1; j < m;j++){
                matrix[i][j] += matrix[i][j-1];
            }
        }

        map<int,int> mp;

        for(int j = 0; j < m;j++){
            for(int k = j; k < m;k++){
                mp.clear();
                mp[0] = 1;
                int sum = 0;
                for(int i = 0; i < n;i++){
                    sum += matrix[i][k];
                    if(j){
                        sum -= matrix[i][j-1];
                    }
                    auto p = mp.lower_bound(sum - target);
                    // if(sum < 0 && target > 0) continue;
                    if(p == mp.end()){
                        mp[sum]++;
                        continue;
                    }
                    if(ans < sum - p->first){
                        // cout<<sum-target<<" "<<p->first<<"\n";
                        // cout<<i<<" "<<j<<" "<<k<<" "<<p->first<<"\n";
                        ans = (sum - p->first);
                    }
                    mp[sum]++;
                }
            }
        }
        return ans;
    }
};
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