764. Largest Plus Sign
In a 2D grid
from (0, 0) to (N-1, N-1), every cell contains a 1
, except those cells in the given list mines
which are 0
. What is the largest axis-aligned plus sign of 1
s contained in the grid? Return the order of the plus sign. If there is none, return 0.
An "axis-aligned plus sign of 1
s of order k" has some center grid[x][y] = 1
along with 4 arms of length k-1
going up, down, left, and right, and made of 1
s. This is demonstrated in the diagrams below. Note that there could be 0
s or 1
s beyond the arms of the plus sign, only the relevant area of the plus sign is checked for 1s.
Examples of Axis-Aligned Plus Signs of Order k:
Order 1: 000 010 000 Order 2: 00000 00100 01110 00100 00000 Order 3: 0000000 0001000 0001000 0111110 0001000 0001000 0000000
Example 1:
Input: N = 5, mines = [[4, 2]] Output: 2 Explanation: 11111 11111 11111 11111 11011 In the above grid, the largest plus sign can only be order 2. One of them is marked in bold.
Example 2:
Input: N = 2, mines = [] Output: 1 Explanation: There is no plus sign of order 2, but there is of order 1.
Example 3:
Input: N = 1, mines = [[0, 0]] Output: 0 Explanation: There is no plus sign, so return 0.
Note:
N
will be an integer in the range[1, 500]
.mines
will have length at most5000
.mines[i]
will be length 2 and consist of integers in the range[0, N-1]
.- (Additionally, programs submitted in C, C++, or C# will be judged with a slightly smaller time limit.)
Rust Solution
struct Solution;
impl Solution {
fn order_of_largest_plus_sign(n: i32, mines: Vec<Vec<i32>>) -> i32 {
let n = n as usize;
let mut grid = vec![vec![1; n]; n];
let mut left = vec![vec![0; n]; n];
let mut top = vec![vec![0; n]; n];
let mut right = vec![vec![0; n]; n];
let mut bottom = vec![vec![0; n]; n];
for mine in mines {
let i = mine[0] as usize;
let j = mine[1] as usize;
grid[i][j] = 0;
}
for i in 0..n {
for j in 0..n {
if grid[i][j] == 1 {
if j > 0 {
left[i][j] = left[i][j - 1] + 1;
} else {
left[i][j] = 1;
}
}
}
}
for j in 0..n {
for i in 0..n {
if grid[i][j] == 1 {
if i > 0 {
top[i][j] = top[i - 1][j] + 1;
} else {
top[i][j] = 1;
}
}
}
}
for i in 0..n {
for j in (0..n).rev() {
if grid[i][j] == 1 {
if j + 1 < n {
right[i][j] = right[i][j + 1] + 1;
} else {
right[i][j] = 1;
}
}
}
}
for j in 0..n {
for i in (0..n).rev() {
if grid[i][j] == 1 {
if i + 1 < n {
bottom[i][j] = bottom[i + 1][j] + 1;
} else {
bottom[i][j] = 1;
}
}
}
}
let mut res = 0;
for i in 0..n {
for j in 0..n {
let mut min = n;
min = min.min(left[i][j]);
min = min.min(right[i][j]);
min = min.min(top[i][j]);
min = min.min(bottom[i][j]);
res = res.max(min);
}
}
res as i32
}
}
#[test]
fn test() {
let n = 5;
let mines = vec_vec_i32![[4, 2]];
let res = 2;
assert_eq!(Solution::order_of_largest_plus_sign(n, mines), res);
let n = 2;
let mines = vec_vec_i32![];
let res = 1;
assert_eq!(Solution::order_of_largest_plus_sign(n, mines), res);
let n = 1;
let mines = vec_vec_i32![[0, 0]];
let res = 0;
assert_eq!(Solution::order_of_largest_plus_sign(n, mines), res);
}
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