<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(), ans = INT_MIN;
        vector <vector <int>> pref(n, vector <int> (m, 0));

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (j == 0)
                    pref[i][j] = matrix[i][j];
                else
                    pref[i][j] = pref[i][j - 1] + matrix[i][j];
            }
        }

        for (int i = 0; i < m; i++) {
            for (int j = 0; j <= i; j++) {
                vector <int> arr(n);
                set <int> s;
                for (int k = 0; k < n; k++) {
                    if (j == 0)
                        arr[k] = pref[k][i];
                    else
                        arr[k] = pref[k][i] - pref[k][j - 1];
                    
                    if (k > 0)
                        arr[k] += arr[k - 1];
                    
                    if (arr[k] <= target)
                        ans = max(ans, arr[k]);
                    
                    auto lb = s.lower_bound(arr[k] - target);
                    if (lb != s.end())
                        ans = max(ans, arr[k] - *lb);
                    s.insert(arr[k]);

                    if (ans == target)
                        return ans;
                }
            }
        }

        return ((ans == INT_MIN) ? -1 : ans);
    }
};
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