<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 dp[5001][4][4][4] = {};
    int numOfWays(int n) {
        return dfs(n, 0, 0, 0);
    }
    int dfs(int n, int a0, int b0, int c0) {
        if (n == 0) return 1;
        if (dp[n][a0][b0][c0] != 0) return dp[n][a0][b0][c0];
        int ans = 0;
        vector<int> colors = {1, 2, 3}; 
        for (int a : colors) {
            if (a == a0) continue; 
            for (int b : colors) {
                if (b == b0 || b == a) continue; 
                for (int c : colors) {
                    if (c == c0 || c == b) continue; 
                    ans += dfs(n - 1, a, b, c);
                    ans %= 1000000007;
                }
            }
        }
        return dp[n][a0][b0][c0] = ans;
    }
};
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