<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 kadane(int arr[], int n, int k) {
        set<int> prefixSums;
        prefixSums.insert(0);  

        int sum = 0;
        int maxSum = INT_MIN;
        
        for (int i = 0; i < n; i++) {
            sum += arr[i];
            // Find the smallest prefix sum which is greater than or equal to (sum - k)
            auto it = prefixSums.lower_bound(sum - k);
            if (it != prefixSums.end()) {
                maxSum = max(maxSum, sum - *it);
            }
            if(maxSum == k) return k;
            prefixSums.insert(sum);
        }
        
        return maxSum;
    }

    int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
        int m = matrix.size(); // row
        int n = matrix[0].size(); // col
        int arr[n];
        int maxi = INT_MIN;
        for (int i = 0; i < m; i++) {
            memset(arr, 0, sizeof(arr));
            for (int j = i; j < m; j++) {
                for (int col = 0; col < n; col++) {
                    arr[col] += matrix[j][col];
                }
                maxi = max(kadane(arr, n, k), maxi);
                if (maxi == k) {
                    return maxi;
                }
            }
        }
        return maxi>k?-1:maxi;
    }
};
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