<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>

class Solution {
public:
    vector<int> prisonAfterNDays(vector<int>& cells, int n) {
        map<int,vector<int>> mp;
        map<vector<int>,int> mp1;
        int i=0;
        while(i<n){
            if(mp1.find(cells)!=mp1.end()){
                break;
            }
            mp1[cells]=i;
            mp[i]=cells;
            vector<int> temp=cells;
            for(int j=0;j<cells.size();j++){
                if(j==0 or j==cells.size()-1){
                    temp[j]=0;
                }else if((cells[j-1]==0 and cells[j+1]==0) or (cells[j-1]==1 and cells[j+1]==1)){
                    temp[j]=1;
                }else{
                    temp[j]=0;
                }
            }
            cells=temp;
            i++;
        }
        if(i==n){
            return cells;
        }
        int k=mp1[cells];
        int diff=i-k;
        n-=i;
        n%=diff;
        k+=n;
        return mp[k];
    }
};
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