<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:
        mod = 10**9 + 7
        types = list()
        for i in range(3):
            for j in range(3):
                for k in range(3):
                    if i != j and j != k:
                        types.append(i * 9 + j * 3 + k)
        type_cnt = len(types)
        related = [[0] * type_cnt for _ in range(type_cnt)]
        for i, ti in enumerate(types):
            x1, x2, x3 = ti // 9, ti // 3 % 3, ti % 3
            for j, tj in enumerate(types):
                y1, y2, y3 = tj // 9, tj // 3 % 3, tj % 3
                if x1 != y1 and x2 != y2 and x3 != y3:
                    related[i][j] = 1
        f = [[0] * type_cnt for _ in range(n + 1)]
        f[1] = [1] * type_cnt
        for i in range(2, n + 1):
            for j in range(type_cnt):
                for k in range(type_cnt):
                    if related[k][j]:
                        f[i][j] += f[i - 1][k]
                        f[i][j] %= mod
        ans = sum(f[n]) % mod
        return ans

        
19300