<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) {

        std::vector<int> rowSum(matrix[0].size());
        for(int i = 0; i < matrix.size(); ++i) {

            fill(rowSum.begin(), rowSum.end(), 0);
            
            for(int row = i; row < matrix.size(); ++row) {

                for(int col = 0; col < matrix[0].size(); ++col) {
                    rowSum[col] += matrix[row][col];
                }

                updateResult(rowSum, k);

                if(result == k) {
                    return result;
                }
            }
        }

        return result;
    }

private:
        int result = INT_MIN;
        void updateResult(std::vector<int>& nums, int k) {

            int sum = 0;
            std::set<int> sortedSum;
            std::set<int>::iterator it;

            sortedSum.insert(0);
            for(int & num : nums) {

                sum += num;
                it = sortedSum.lower_bound(sum - k);
                
                if(it != sortedSum.end()) {
                    result = std::max(result, sum - *it);
                }

                sortedSum.insert(sum);
            }
        }


};
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