Prison Cells After N Days

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 = 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:

1. cells.length == 8

2. cells[i] is in {0, 1}

3. 1 <= N <= 10^9

class Solution {
    public int[] prisonAfterNDays(int[] cells, int N) {
        if (cells == null || cells.length == 0 || N <= 0)
            return cells;
        boolean hasCycle = false;
        int cycle = 0;
        HashSet<String> set = new HashSet<>();
        for (int i = 0; i < N; i++) {
            int[] next = nextDay(cells);
            String key = Arrays.toString(next);
            if (!set.contains(key)) { // store cell state
                set.add(key);
                cycle++;
            } else { // hit a cycle
                hasCycle = true;
                break;
            }
            cells = next;
        }
        if (hasCycle) {
            N %= cycle;
            for (int i = 0; i < N; i++) {
                cells = nextDay(cells);
            }
        }
        return cells;
    }

    private int[] nextDay(int[] cells) {
        int[] tmp = new int[cells.length];
        for (int i = 1; i < cells.length - 1; i++) {
            tmp[i] = cells[i - 1] == cells[i + 1] ? 1 : 0;
        }
        return tmp;
    }
}