# Leetcode 957. Prison Cells After N Days

[medium][Amazon]

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.

You are given an integer array `cells`

where `cells[i] == 1`

if the `ith`

cell is occupied and `cells[i] == 0`

if the `ith`

cell is vacant, and you are given an integer `n`

.

Return the state of the prison after `n`

days (i.e., `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]

**Constraints:**

`cells.length == 8`

`cells[i]`

is either`0`

or`1`

.`1 <= n <= 109`

**[Gary- brute force- Java]**

- can’t pass all the test case — TLE
- need to find pattern

[Second] — using hashmap to store all the possible string pattern

- Trick part: n %= map.get(gen) — n