3rd-July-Prison-Cells-After-N-Days.cpp

Problem :
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

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Solution:
class Solution {
public:
    vector<int> prisonAfterNDays(vector<int>& cells, int N) {
        N= N%14==0 ? 14 : N%14;    //The pattern will repeat after every 14 days. This mean 1st day and 15th day will have same cells so we dont have use all n days.
       
        for(int k=0;k<N;k++)
        {
            vector<int> temp(cells.size());
            for(int i=1;i<cells.size()-1;i++){
                if(cells[i-1] ^ cells[i+1]==0){
                    temp[i]=1;
                }
                else{
                    temp[i]=0;
                }
            }
            
           cells=temp;
          
        }
        
        return cells;
        
    }
};
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Done by
Md Mobbasher Ansari


258 / 258 test cases passed.
Status: Accepted
Runtime: 0 ms
Memory Usage: 11.9 MB