<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>

int p = 1e9+7;
vector<vector<vector<vector<int>>>> dp;
class Solution {
public:
    int rec(int i, int a, int b ,int c, int n){
        if(i==n) return 1;
        if(dp[i][a][b][c]!=-1) return dp[i][a][b][c];
        int ans = 0;
        for(int x=0;x<3;x++)
            for(int y=0;y<3;y++)
                for(int z=0;z<3;z++){
                    if(x==y or y==z or x==a or y==b or z==c) continue;
                    ans=(ans+rec(i+1,x,y,z,n))%p;
                }
        return dp[i][a][b][c] = ans%p;
    }
    int numOfWays(int n) {
        dp.assign(n, 
            vector<vector<vector<int>>>(4, 
                vector<vector<int>>(4, 
                    vector<int>(4,-1)
                )
            )
        );
        return rec(0,3,3,3,n);
    }
};
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