<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 v[3]={1,2,3};
 int check( int a, int b)
 {
        while( a && b)
    {
        if( a%10==b%10)
        {
            return 0;
        }
        a=a/10;
        b=b/10;
    }
    return 1;
 }
 int mod=1e9+7;
int func( int a, int b, vector<vector<int>> & dp, vector<int> & v)
{
    if( a==dp.size())
    {
        return 1;
    }
    if( dp[a][b]!=-1)
    {
        return dp[a][b];
    }
    int cnt=0;
    for( int j=0; j<v.size() ; j++)
    {
        if( check( v[j], b))
        {
            cnt= (cnt+(func( a+1,v[j], dp, v))%mod)%mod;
        }
          
    }
    dp[a][b]=cnt;
    return cnt;
   

}
    int numOfWays(int n) {
        set<int> s;
        for( int j=0; j<3; j++)
        {
            for ( int k=0; k<3; k++)
            {
                for( int l=0; l<3; l++)
                {
                    if( j!=k && k!=l)
                    {
                        int aa=(100*v[j])+ ( 10*v[k])+(v[l]);
                        s.insert( aa);
                    }

                }
            }

        }
        vector<vector<int>>dp(n, vector<int> ( 350, -1));
        vector<int> v;
        for( auto it: s)
        {
            v.push_back( it);
        }
       
       return func( 0,0,dp, v);
        
    }
};
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