<p>You have a <code>grid</code> of size <code>n x 3</code> and you want to paint each cell of the grid with exactly one of the three colors: <strong>Red</strong>, <strong>Yellow,</strong> or <strong>Green</strong> while making sure that no two adjacent cells have the same color (i.e., no two cells that share vertical or horizontal sides have the same color).</p>

<p>Given <code>n</code> the number of rows of the grid, return <em>the number of ways</em> you can paint this <code>grid</code>. As the answer may grow large, the answer <strong>must be</strong> computed modulo <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/03/26/e1.png" style="width: 400px; height: 257px;" />
<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 12
<strong>Explanation:</strong> There are 12 possible way to paint the grid as shown.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 5000
<strong>Output:</strong> 30228214
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>1 &lt;= n &lt;= 5000</code></li>
</ul>

class Solution {
public:
int dp[1024][5001][4];
int m=1e9+7;
int fun(int c,int i,int j,int n)
{ 
    long long ans=0;
   
    if(i==n)return 1;
    if(j==3)
  {  return fun(c,i+1,0,n);}
    if(dp[c][i][j]!=-1){
        return dp[c][i][j];}
   
   
    for(int l=0;l<3;l++)
    {
         int c1=c;
      
        if(((c1&(int)pow(2,2*3+l))==0))
        {
            if(((j-1)>=0)&&(c1&(int)pow(2,l))==0)
           { 
            c1*=8;
        if(c1&(int)(pow(2,9)))
        c1-=pow(2,9);
        if(c1&((int)pow(2,10)))
        c1-=pow(2,10);
        if(c1&((int)pow(2,11)))
        c1-=pow(2,11);
      
        
        ans+=fun(c1|(int)pow(2,l),i,j+1,n);
        ans%=m;}
        else if(j==0)
        {
            
            c1*=8;
        if(c1&(int)(pow(2,9)))
        c1-=pow(2,9);
        if(c1&((int)pow(2,10)))
        c1-=pow(2,10);
        if(c1&((int)pow(2,11)))
        c1-=pow(2,11);
        
            
        
        ans+=fun(c1|(int)pow(2,l),i,j+1,n); 
         ans%=m;   
        }

        }
    }
    return dp[c][i][j]=ans;

}
    int numOfWays(int n) {
        memset(dp,-1,sizeof(dp));
        return fun(0,0,0,n);
    }
};
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