LeetCode [1292] Maximum Side Length of a Square with Sum Less than or Equal to Threshold

 1292. Maximum Side Length of a Square with Sum Less than or Equal to Threshold

Medium

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

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class Solution { public int maxSideLength(int[][] mat, int threshold) { int n = mat.length; int m = mat[0].length; int[][] sums = new int[n + 1][m + 1]; int max = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { sums[i + 1][j + 1] = sums[i + 1][j] + sums[i][j + 1] - sums[i][j] + mat[i][j]; if (i - max >= 0 && j - max >= 0 && sums[i + 1][j + 1] - sums[i - max][j + 1] - sums[i + 1][j - max] + sums[i - max][j - max] <= threshold ) { max += 1; } } } return max; } }