# 562. Longest Line of Consecutive One in Matrix

## 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

• 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;
}
}

• class Solution {
public:
int longestLine(vector<vector<int>>& mat) {
int m = mat.size(), n = mat[0].size();
vector<vector<int>> a(m + 2, vector<int>(n + 2));
vector<vector<int>> b(m + 2, vector<int>(n + 2));
vector<vector<int>> c(m + 2, vector<int>(n + 2));
vector<vector<int>> d(m + 2, vector<int>(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]) {
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, max(a[i][j], max(b[i][j], max(c[i][j], d[i][j]))));
}
}
}
return ans;
}
};

• 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):
v = mat[i - 1][j - 1]
if v:
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


• func longestLine(mat [][]int) (ans int) {
m, n := len(mat), len(mat[0])
f := make([][][4]int, m+2)
for i := range f {
f[i] = make([][4]int, n+2)
}
for i := 1; i <= m; i++ {
for j := 1; j <= n; j++ {
if mat[i-1][j-1] == 1 {
f[i][j][0] = f[i-1][j][0] + 1
f[i][j][1] = f[i][j-1][1] + 1
f[i][j][2] = f[i-1][j-1][2] + 1
f[i][j][3] = f[i-1][j+1][3] + 1
for _, v := range f[i][j] {
if ans < v {
ans = v
}
}
}
}
}
return
}