<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:
    def numOfWays(self, n: int) -> int:
        prev = [set() for i in range(27)]
        valid_s = []
        for i in range(27):
            a, b, c = i % 3, i // 3 % 3, i // 9 % 3
            if a == b or b == c:
                continue
            else:
                valid_s.append(i)
        for i in valid_s:
            a, b, c = i % 3, i // 3 % 3, i // 9 % 3
            for j in valid_s:
                x, y, z = j % 3, j // 3 % 3, j // 9 % 3
                if a == x or b == y or c == z:
                    continue
                else:
                    prev[i].add(j)
                    prev[j].add(i)
        f = [[0] * 27 for i in range(n)]
        for s in valid_s:
            f[0][s] = 1
        mod = int(1e9 + 7)
        for i in range(1, n):
            for s in valid_s:
                for p in prev[s]:
                    f[i][s] = (f[i][s] + f[i - 1][p]) % mod
        ans = 0
        for s in valid_s:
            ans = (ans + f[n - 1][s]) % mod
        return ans
        
19800