# Prison Cells After N Days Problem

## Description

LeetCode Problem 957.

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.

You are given an integer array cells where cells[i] == 1 if the i^th cell is occupied and cells[i] == 0 if the i^th cell is vacant, and you are given an integer n.

Return the state of the prison after n days (i.e., n such changes described above).

Example 1:

``````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:

``````1
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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]
``````

Constraints:

• cells.length == 8
• cells[i]is either 0 or 1.
• 1 <= n <= 10^9

## Sample C++ Code

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class Solution {
public:
vector<int> prisonAfterNDays(vector<int>& cells, int N) {
int n = cells.size(), cycle = 0;
vector<int> cur(n, 0), direct;
while (N-- > 0) {
for (int i = 1; i < n - 1; ++i)
cur[i] = cells[i - 1] == cells[i + 1];
if (direct.empty())
direct = cur;
else if (direct == cur)
N %= cycle;

++cycle;
cells = cur;
}
return cur;
}
};
``````