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:

1. cells.length == 8
2. cells[i] is in {0, 1}
3. 1 <= N <= 10^9

# Solution: Simulation

Simulate the process, since there must be loops, record the last day when a state occurred.

Time complexity: O(2^8)
Space complexity: O(2^8)

## C++

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