Leetcode_5285_Maximum Side Length of a Square with Sum Less than or Equal to Threshold

[lihe]

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 result = 0;

        for(int i = 0; i < mat.length; i++){
            for(int j = 0; j < mat[0].length; j++){
                if(j != 0){
                    mat[i][j] += mat[i][j - 1];
                }
                int len = 0;
                int maxlen = Math.min(i, j) + 1;
                while (len < maxlen){
                    int area = 0;
                    for(int k = 0; k < len + 1; k++){
                        int prefix = j - len - 1 < 0 ? 0 : mat[i - k][j - len - 1];
                        area += mat[i - k][j] - prefix;
                    }
                    if(area > threshold) {
                        break;
                    }
                    len++;
                }
                result =  len > result ? len : result;
            }
        }
        return result;
    }
}