<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:
        dp = [[0] *27 for _ in range(n)]
        def three(x):
            ans = []
            while x:
                ans.append(x % 3)
                x //= 3
            while len(ans) < 3:
                ans.append(0)
            return ans
        def row_judge(x):
            return all(i != j for i, j in pairwise(three(x)))
        def check_valid(x, y):
            return all(i != j for i, j in zip(three(x), three(y)))
        one_row = [i for i in range(27) if row_judge(i)]
        valid = [[] for _ in range(27)]
        for i in one_row:
            for j in one_row:
                if check_valid(i, j):
                    valid[i].append(j)

        mod = 10 ** 9 + 7
        for i in range(n):
            for j in one_row:
                if i == 0: dp[i][j] = 1
                else: 
                    for k in valid[j]: 
                        dp[i][j] = (dp[i][j] + dp[i - 1][k]) % mod
        return sum(dp[n - 1]) % mod
                
        

19800