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:

- If a cell has two adjacent neighbors that are both occupied or both vacant, then the cell becomes occupied.
- Otherwise, it becomes vacant.

(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 = 7Output:[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 = 1000000000Output:[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);
}
```