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 1s contained in the grid? Return the order of the plus sign. If there is none, return 0.
An "axis-aligned plus sign of 1s 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 1s. This is demonstrated in the diagrams below. Note that there could be 0s or 1s 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:
Nwill be an integer in the range[1, 500].mineswill 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);
}
Having problems with this solution? Click here to submit an issue on github.