<p>There are <code>8</code> prison cells in a row and each cell is either occupied or vacant.</p>

<p>Each day, whether the cell is occupied or vacant changes according to the following rules:</p>

<ul>
	<li>If a cell has two adjacent neighbors that are both occupied or both vacant, then the cell becomes occupied.</li>
	<li>Otherwise, it becomes vacant.</li>
</ul>

<p><strong>Note</strong> that because the prison is a row, the first and the last cells in the row can&#39;t have two adjacent neighbors.</p>

<p>You are given an integer array <code>cells</code> where <code>cells[i] == 1</code> if the <code>i<sup>th</sup></code> cell is occupied and <code>cells[i] == 0</code> if the <code>i<sup>th</sup></code> cell is vacant, and you are given an integer <code>n</code>.</p>

<p>Return the state of the prison after <code>n</code> days (i.e., <code>n</code> such changes described above).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> cells = [0,1,0,1,1,0,0,1], n = 7
<strong>Output:</strong> [0,0,1,1,0,0,0,0]
<strong>Explanation:</strong> 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]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> cells = [1,0,0,1,0,0,1,0], n = 1000000000
<strong>Output:</strong> [0,0,1,1,1,1,1,0]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>cells.length == 8</code></li>
	<li><code>cells[i]</code>&nbsp;is either <code>0</code> or <code>1</code>.</li>
	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def prisonAfterNDays(self, cells: List[int], n: int) -> List[int]:
        table = [([0] * 8) for _ in range(14)]
        prev = cells
        table[0] = cells
        for i in range(1, 28):
            cur = [0] * 8
            for j in range(1, 7):
                if prev[j - 1] + prev[j + 1] == 2 or prev[j - 1] + prev[j + 1] == 0:
                    cur[j] = 1
            prev = cur
            table[i % 14] = prev
        return table[n % 14]
16600