Day-03-Prison Cells After N Days.cpp

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 = 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:

Input: cells = [1,0,0,1,0,0,1,0], N = 1000000000
Output: [0,0,1,1,1,1,1,0]


Note:

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








class Solution {
public:
    vector<int> prisonAfterNDays(vector<int>& cells, int N) {
        int n=cells.size();
        vector<int> res=cells;

        bool first=true;
        if(N>14) N=(N%14)+14;
        while(N--){
            for(int i=1;i<n-1;i++){
                if(cells[i-1]==1 && cells[i+1]==1) res[i]=1;
                else if(cells[i-1]==0 && cells[i+1]==0) res[i]=1;
                else res[i]=0;
            }
            if(first){
                res[0]=0;
                res[n-1]=0;
                first=false;
            }
            cells=res;
        }

        return res;
    }
};