<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>

vector<string> v= {"121","123","131","132","212","213","231","232","312","313","321","323"};

const int MOD= 1000000007;
class Solution {

    bool check(int i1, int prev)
    {
        string s1= v[i1];
        string s2= v[prev];
        for(int i=0;i<3;i++)
        {
            if(s1[i]==s2[i])
            {
                return false;
            }
        }
        return true;
    }

    int f(int n , int prev, vector<string> &v)
    {
        if(n==0)
        {
            return 1;
        }
        int ans=0;
        for(int i=0;i<12;i++)
        {
            if(check(prev,i))
            {
                ans= ans+f(n-1,i,v);
            }
        }
        return ans;
    }
public:
    int numOfWays(int n) {
        int ans=0;

        vector<vector<int>> dp (n+1, vector<int> (12,0));
        for(int i=0;i<12;i++)
        {
            dp[0][i]=1;
        }

        
            for(int k=1;k<=n-1;k++)
            {
                for(int prev=0;prev<12;prev++)
                {
                    int ans=0;
                    for(int j=0;j<12;j++)
                    {
                        if(check(prev,j))
                        {
                            ans= (ans%MOD+dp[k-1][j]%MOD)%MOD;
                        }
                    }
                    dp[k][prev]= ans%MOD;   
                }
            }

            for(int i=0;i<12;i++)
            {
                ans= (ans%MOD+ dp[n-1][i]%MOD)%MOD;
            }
            
        return ans%MOD;
    }
};
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