<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 solve(int i, int j, int n, vector<vector<int>> &v, vector<vector<int>> &dp)
    {
        int mod=1e9+7;
        if(dp[i][j]!=-1)
        return dp[i][j];
        if(i==n)
        return dp[i][j]=1;
        else if(j==12)
        {
            long long ans=0;
            for(int k=0;k<12;k++)
            {
                ans=(ans+solve(i+1,k,n,v,dp))%mod;
            }
            return dp[i][j]=ans%mod;
        }
        else{
            long long ans=0;
            for(int k=0;k<12;k++)
            {
                bool flag=true;
                for(int x=0;x<3;x++)
                {
                    if(v[j][x]==v[k][x])
                    {
                        flag=false;
                        break;
                    }
                }
                if(flag)
                {
                    ans=(ans+solve(i+1,k,n,v,dp));
                }
            }
            return dp[i][j]=ans%mod;
        }
    }
    
    int numOfWays(int n) {
        vector<vector<int>> v;
        v.push_back({0,1,0});
        v.push_back({0,2,0});
        v.push_back({1,0,1});
        v.push_back({1,2,1});
        v.push_back({2,1,2});
        v.push_back({2,0,2});
        v.push_back({2,1,0});
        v.push_back({2,0,1});
        v.push_back({1,0,2});
        v.push_back({1,2,0});
        v.push_back({0,1,2});
        v.push_back({0,2,1});
        vector<vector<int>> dp(n+1,vector<int>(13,-1));
        return solve(0,12,n,v,dp);
    }
};
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