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:

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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:

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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

Solution:

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class Solution {
    public int[] prisonAfterNDays(int[] cells, int N) {
        N = (N - 1) % 14 + 1;
        while (N > 0) {
            cells = nextDay(cells);
             N--;
        }
        return cells;
    }

    int[] nextDay(int[] cells) {
        int[] nextDay = new int[cells.length];
        for (int i = 1; i < cells.length - 1; i++) {
             if (cells[i - 1] == cells[i + 1]) {
                 nextDay[i] = 1;
             } else {
                 nextDay[i] = 0;
             }
         }
         return nextDay;
    }
}