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

LeetCode-1292. 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 = 4Output: 2Explanation: 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 = 1Output: 0

Example 3:

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

Example 4:

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

Constraints:

​​1 <= m, n <= 300​​​​m == mat.length​​​​n == mat[i].length​​​​0 <= mat[i][j] <= 10000​​​​0 <= threshold <= 10^5​​

​​题解：​​

从大到小枚举所有正方形。

class Solution {public: int maxSideLength(vector>& mat, int threshold) { int m = mat.size(), n = mat[0].size(); vector> sum(m + 1, vector(n + 1, 0)); for (int i = 1; i <= m; i++) { for (int j = 1; j <= n; j++) { sum[i][j] = mat[i - 1][j - 1] + sum[i][j - 1] + sum[i - 1][j] - sum[i - 1][j - 1]; } } int k = min(m, n); while (k > 0) { for (int i = 0; i <= m - k; i++) { for (int j = 0; j <= n - k; j++) { if (sum[i + k][j + k] + sum[i][j] - sum[i + k][j] - sum[i][j + k] <= threshold) { return k; } } } k--; } return 0; }};

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