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562. Longest Line of Consecutive One in Matrix πŸ”’

Description

Given an m x n binary matrix mat, return the length of the longest line of consecutive one in the matrix.

The line could be horizontal, vertical, diagonal, or anti-diagonal.

 

Example 1:

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

Example 2:

Input: mat = [[1,1,1,1],[0,1,1,0],[0,0,0,1]]
Output: 4

 

Constraints:

  • m == mat.length
  • n == mat[i].length
  • 1 <= m, n <= 104
  • 1 <= m * n <= 104
  • mat[i][j] is either 0 or 1.

Solutions

Solution 1

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class Solution:
    def longestLine(self, mat: List[List[int]]) -> int:
        m, n = len(mat), len(mat[0])
        a = [[0] * (n + 2) for _ in range(m + 2)]
        b = [[0] * (n + 2) for _ in range(m + 2)]
        c = [[0] * (n + 2) for _ in range(m + 2)]
        d = [[0] * (n + 2) for _ in range(m + 2)]
        ans = 0
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if mat[i - 1][j - 1]:
                    a[i][j] = a[i - 1][j] + 1
                    b[i][j] = b[i][j - 1] + 1
                    c[i][j] = c[i - 1][j - 1] + 1
                    d[i][j] = d[i - 1][j + 1] + 1
                    ans = max(ans, a[i][j], b[i][j], c[i][j], d[i][j])
        return ans
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class Solution {
    public int longestLine(int[][] mat) {
        int m = mat.length, n = mat[0].length;
        int[][] a = new int[m + 2][n + 2];
        int[][] b = new int[m + 2][n + 2];
        int[][] c = new int[m + 2][n + 2];
        int[][] d = new int[m + 2][n + 2];
        int ans = 0;
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (mat[i - 1][j - 1] == 1) {
                    a[i][j] = a[i - 1][j] + 1;
                    b[i][j] = b[i][j - 1] + 1;
                    c[i][j] = c[i - 1][j - 1] + 1;
                    d[i][j] = d[i - 1][j + 1] + 1;
                    ans = max(ans, a[i][j], b[i][j], c[i][j], d[i][j]);
                }
            }
        }
        return ans;
    }

    private int max(int... arr) {
        int ans = 0;
        for (int v : arr) {
            ans = Math.max(ans, v);
        }
        return ans;
    }
}
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