PRISON CELLS AFTER N DAYS

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:

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]

Note:

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

Solution:

class Solution {
public:
    
vector<int> prisonAfterNDays(vector<int>& cells, int N) {
    vector<int> ans(8, 0);
    int count  = N % 14;
    if (count == 0) count = 14;
    for(int i = 0; i < count; i++){   
        for (int j = 1; j < 7; j++)
            ans[j] = ((cells[j-1] ^ cells[j+1])!=0) ? 0:1;             
        cells = ans; 
    }
    return ans;			
}
};