Maximum Side Length of a Square with Sum Less than or Equal to Threshold

Given a m x n matrix mat and an integer threshold. Return the maximum side-length of a square with a sum less than or equal to threshold or return 0 if there is no such square.

Example 1:

Input: mat = [[1,1,3,2,4,3,2],[1,1,3,2,4,3,2],[1,1,3,2,4,3,2]], threshold = 4
Output: 2
Explanation: The maximum side length of square with sum less than 4 is 2 as shown.

Example 2:

Input: mat = [[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2]], threshold = 1
Output: 0

Example 3:

Input: mat = [[1,1,1,1],[1,0,0,0],[1,0,0,0],[1,0,0,0]], threshold = 6
Output: 3

Example 4:

Input: mat = [[18,70],[61,1],[25,85],[14,40],[11,96],[97,96],[63,45]], threshold = 40184
Output: 2

Constraints:

• 1 <= m, n <= 300

• m == mat.length

• n == mat[i].length

• 0 <= mat[i][j] <= 10000

• 0 <= threshold <= 10^5

class Solution {
    public int maxSideLength(int[][] mat, int threshold) {
        int n = mat.length;
        int m = mat[0].length;
        int[][] prefixSum = new int[n + 1][m + 1];
        int max = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                prefixSum[i + 1][j + 1] = prefixSum[i + 1][j] + prefixSum[i][j + 1] - prefixSum[i][j] + mat[i][j];
                // If we consider a square with bottom-right at (i,j)
                // now if this square's edge length needs to be > max, for it to be a better
                // answer
                if (i - max >= 0 && j - max >= 0 && prefixSum[i + 1][j + 1] - prefixSum[i - max][j + 1]
                        - prefixSum[i + 1][j - max] + prefixSum[i - max][j - max] <= threshold)
                    max += 1;
            }
        }
        return max;
    }
}