<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 M= 1e9+7;
    int t[5002][4][4][4][4][4];
    int f(int i, int j,int p,int a, int b, int c,int n){
       // int n= v.size();
        if(j==3){
            return (f(i+1,0,0,a,b,c,n))%M;
        }
        if(i>=n ){
            return 1;
        }
        if(t[i][j][p][a][b][c] !=-1){
            return t[i][j][p][a][b][c];
        }
        int ans=0;
        
        for(int k=1; k<=3; k++){
            if(j==0){
                 if(p!=k && a!=k){
                    ans= (ans+f(i,j+1,k,k,b,c,n))%M;
                 }
            }
             if(j==1){
                 if(p!=k && b!=k){
                    ans= (ans+f(i,j+1,k,a,k,c,n))%M;
                 }
            }
             if(j==2){
                 if(p!=k && c!=k){
                    ans= (ans+f(i,j+1,k,a,b,k,n))%M;
                 }
            }
           
        }
       
        return t[i][j][p][a][b][c]=ans;


    }
    int numOfWays(int n) {
       // vector<vector<int>>v(n,vector<int>(3,0));
      //  vector<vector<int>>v(n,vector<int>(3,0));
      memset(t,-1,sizeof(t));
        return f(0,0,0,0,0,0,n);
    }
};
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