<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 max_sum(vector<int>&nums ,int k){
        int ans = INT_MIN;
        set<int>st;
        int sum = 0;
        for(int i = 0;i<nums.size();i+=1){
            sum += nums[i];
            if(sum<=k)ans = max(ans ,sum);
            auto it = st.lower_bound(sum-k);
            if(it != st.end()){
                ans = max(ans , sum - *it);
            }
            st.insert(sum);
        }
        return ans;
    }
    int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
            int ans =INT_MIN;
            int n = matrix.size();
            int m = matrix[0].size();
            for(int i =0;i<n;i+=1){
                vector<int>temp(m,0);
                for(int j = i;j<n;j+=1){
                    for(int k = 0;k<m;k+=1){
                        temp[k] += matrix[j][k];
                    }
                    ans = max(ans , max_sum(temp , k));
                }
            }
            return ans;
    }
};
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