<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:
    vector<string> colors = {"RYR","YRY","GRY",
                            "RYG","YRG","GRG",
                            "RGR","YGR","GYR",
                            "RGY","YGY","GYG"};
    int mod = 1e9 + 7;
    vector<vector<int>> dp;
    int get(int n,int pre)
    {
        if(n==0)
            return 1;
        
        if(dp[n][pre+1]!=-1)
            return dp[n][pre+1];
        int res = 0;
        if(pre == -1)
        {
            for(int i=0;i<colors.size();i++)
            {
                res = (res+get(n-1,i))%mod;
            }
        }
        else
        {
            for(int i=0;i<colors.size();i++)
            {
                if(colors[pre][0]!=colors[i][0] && colors[pre][1]!=colors[i][1]  && colors[pre][2]!=colors[i][2] )
                    res = (res+get(n-1,i))%mod;
            }
        }
        return dp[n][pre+1] = res;
    }
    int numOfWays(int n) {
        vector<vector<int>> temp(n+1,vector<int>(13,-1));
        dp = temp;
        return get(n,-1);
    }
};
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