<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>

import numpy as np
class Solution:
    def numOfWays(self, n: int) -> int:
        d = np.zeros((n, 12))
        d[0, :] = 1
        nodes = np.array([
            [0, 1, 0],
            [0, 1, 2],
            [0, 2, 0],
            [0, 2, 1],
            [1, 0, 1],
            [1, 0, 2],
            [1, 2, 1],
            [1, 2, 0],
            [2, 0, 1],
            [2, 0, 2],
            [2, 1, 0],
            [2, 1, 2]
        ])
        edges = {}
        for i in range(12):
            for j in range(12):
                if np.min(np.abs(nodes[i] - nodes[j])) == 0:
                    continue
                if i in edges.keys():
                    edges[i].append(j)
                else:
                    edges[i] = [j]
        print (edges)
        for i in range(1, n):
            for j in range(12):
                for k in edges[j]:
                    d[i,j] = (d[i,j] + d[i-1, k])  % (10**9+7)
                d[i, j] = d[i, j] % (10**9+7)
        return int(np.sum(d[n-1, :])) % (10**9+7)
29700