<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 func(vector<int>& temp, int k) {
        map<int, int> mp;
        int sum = 0;
        int maxi = -1e9;
        mp[0] = 1; 
        for (int i = 0; i < temp.size(); i++) {
            sum += temp[i];
            int target = sum - k;
            auto it = mp.lower_bound(target);
            if (it != mp.end()) {
                maxi = max(maxi, sum - it->first);
            }
            mp[sum]++;
        }
        return maxi;
    }

    int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
        int res = -1e9;
        int rows = matrix.size();
        int cols = matrix[0].size();
        for (int left = 0; left < cols; left++) {
            vector<int> temp(rows, 0);

            for (int right = left; right < cols; right++) {
                for (int i = 0; i < rows; i++) {
                    temp[i] += matrix[i][right];
                }
                res = max(res, func(temp, k));
            }
        }

        return res;
    }
};

201650