There are 8 prison cells in a row, and each cell is either occupied or vacant.
Each day, whether the cell is occupied or vacant changes according to the following rules:
(Note that because the prison is a row, the first and the last cells in the row can't have two adjacent neighbors.)
We describe the current state of the prison in the following way: cells[i] == 1
if the i
-th cell is occupied, else cells[i] == 0
.
Given the initial state of the prison, return the state of the prison after N
days (and N
such changes described above.)
Example 1:
Input: cells = [0,1,0,1,1,0,0,1], N = 7 Output: [0,0,1,1,0,0,0,0] Explanation: The following table summarizes the state of the prison on each day: Day 0: [0, 1, 0, 1, 1, 0, 0, 1] Day 1: [0, 1, 1, 0, 0, 0, 0, 0] Day 2: [0, 0, 0, 0, 1, 1, 1, 0] Day 3: [0, 1, 1, 0, 0, 1, 0, 0] Day 4: [0, 0, 0, 0, 0, 1, 0, 0] Day 5: [0, 1, 1, 1, 0, 1, 0, 0] Day 6: [0, 0, 1, 0, 1, 1, 0, 0] Day 7: [0, 0, 1, 1, 0, 0, 0, 0]
Example 2:
Input: cells = [1,0,0,1,0,0,1,0], N = 1000000000 Output: [0,0,1,1,1,1,1,0]
Note:
cells.length == 8
cells[i]
is in {0, 1}
1 <= N <= 10^9
struct Solution;
use std::collections::HashMap;
impl Solution {
fn prison_after_n_days(mut cells: Vec<i32>, mut n: i32) -> Vec<i32> {
let mut hm: HashMap<Vec<i32>, i32> = HashMap::new();
while n > 0 {
hm.insert(cells.to_vec(), n);
let mut next = vec![0; 8];
for i in 1..7 {
next[i] = 1 - (cells[i - 1] ^ cells[i + 1]);
}
cells = next;
n -= 1;
if let Some(m) = hm.get(&cells) {
n %= m - n;
}
}
cells
}
}
#[test]
fn test() {
let cells = vec![0, 1, 0, 1, 1, 0, 0, 1];
let n = 7;
let res = vec![0, 0, 1, 1, 0, 0, 0, 0];
assert_eq!(Solution::prison_after_n_days(cells, n), res);
let cells = vec![1, 0, 0, 1, 0, 0, 1, 0];
let n = 1_000_000_000;
let res = vec![0, 0, 1, 1, 1, 1, 1, 0];
assert_eq!(Solution::prison_after_n_days(cells, n), res);
}