<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 MOD=1000000007;
    vector<vector<vector<vector<int>>>> dp;
    int N;
    int rec(int level, int col1, int col2, int col3){
        if(level==N){
            return 1;
        }
        if(dp[level][col1][col2][col3]!=-1){
            return dp[level][col1][col2][col3];
        }

        int ans=0;

        for(int i=0;i<3;i++){
            for(int j=0;j<3;j++){
                for(int k=0;k<3;k++){
                    if(i==col1 || j==col2 || k==col3 || i==j || j==k){
                        continue;
                    }
                    else{
                        ans=(ans+rec(level+1,i,j,k))%MOD;
                    }
                }
            }
        }

        return dp[level][col1][col2][col3]=ans;
    }
    int numOfWays(int n) {
        N=n;
        dp.assign(n,vector<vector<vector<int>>>(3,vector<vector<int>>(3,vector<int>(3,-1))));

        int final=0;

        for(int i=0;i<3;i++){
            for(int j=0;j<3;j++){
                for(int k=0;k<3;k++){
                    if(i==j || j==k){
                        continue;
                    }
                    else{
                        final=(final+rec(1,i,j,k))%MOD;
                    }
                }
            }
        }
        return final;
    }

};
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