<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 {
private:
    int len;
    int MOD = 1e9+7;
    int dp[50005][4][4][4] = {0};
    int solver(int n, int c1, int c2, int c3){
        if(n == len){
            return 1;
        }
        if(dp[n][c1][c2][c3]) return dp[n][c1][c2][c3];
        int ans = 0;
        int newc1 = 0, newc2 = 0, newc3 = 0;
        for(int i = 1; i<=3;i++){
            if(i == c1) continue;
            newc1 = i;
            for(int j = 1; j<=3; j++){
                if(newc1 == j || j == c2) continue;
                newc2 = j;
                for(int k = 1; k<=3;k++){
                    if(k == newc2 || k == c3) continue;
                    newc3 = k;
                    ans= (ans + solver(n+1, newc1, newc2, newc3))%MOD;
                }
            }
        }
        return dp[n][c1][c2][c3] = ans%MOD;
    }
public:
    int numOfWays(int n) {
        len = n;
        memset(dp, 0,sizeof(dp[0][0][0][0]) * (n+1)*4*4*4);
        int ans = solver(0, 0, 0, 0);
        return ans;

    }
};
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