<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 solve(vector<int>& temp, int k){
        set<int> s;
        int ans = INT_MIN;
        int acc_sum = 0;
        // a + b = acc_sum
        // a >= acc_sum - k
        for(auto num: temp){
            acc_sum += num;
            if (acc_sum <= k) {
                ans = max(acc_sum,ans);
            }
            auto it = s.lower_bound(acc_sum - k);
            if (it != s.end()){
                ans = max(ans, acc_sum - *it);
            }
            s.insert(acc_sum);
        }
        return ans;
    }

    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];
            }
        }

        int ans = INT_MIN;
        for(int i = 0; i<m; i++){
            for(int j = i; j<m; j++){
                vector<int> temp(n);
                for(int k = 0; k<n; k++){
                    temp[k] = (i!=0) ? matrix[j][k] - matrix[i-1][k] : matrix[j][k];
                }
                ans = max(ans, solve(temp,k));
            }
        }
        return ans;
    }
};
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