<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>


#include <vector>
#include <unordered_map>



class Solution {
public:
    std::vector<int> prisonAfterNDays(std::vector<int>& cells, int n) {
		int curr = 0;
		int next = 0;
		std::vector<int> result;
		std::unordered_map<int, int> day_to_cell;
		std::unordered_map<int, int> cell_to_day;

		for (int i = 0; i < 8; ++i)
			if (cells[i])
				curr |= 1 << i;

		day_to_cell[0] = curr;
		cell_to_day[curr] = 0;

		for (int i = 1; i <= n; ++i) {
			next = 0;

			for (int j = 1; j < 7; ++j)
				if (!(((curr >> (j - 1)) & 1) ^ ((curr >> (j + 1)) & 1)))
					next |= 1 << j;

			curr = next;

			if (cell_to_day.count(curr)) {
				int start = cell_to_day[curr];
				int rotate = i - start;

				curr = day_to_cell[start + (n - start) % rotate];

				break;
			}

			day_to_cell[i] = curr;
			cell_to_day[curr] = i;
		}

		for (int i = 0; i < 8; ++i)
			result.emplace_back((curr & (1 << i)) ? 1 : 0);

		return result;
    }
};
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