<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:
        '''
        
        0 1 2     0 1 2     0 1 2 
        1 0 2     2 0 1     1 2 0
        
        0 1 2     0 1 2
        1 0 1     1 2 1
        
        
        
        
        1 0 1   1 0 1   1 0 1   1 0 1    1 0 1  
        2 1 0   2 1 2   0 1 0   0 2 0    0 1 2  
        
        
       
      
        
        
        
        
        what we know: n = 1: 12 options
        n = 2: 
        
        12 -> 4
        '''
        MOD = 10 ** 9 + 7
        
        @cache
        def calc(index, a1, a2, a3):
            if index == n:
                return 1
            
            total = 0
            for b1, b2, b3 in options:
                if a1 != b1 and a2 != b2 and a3 != b3:
                    total += calc(index + 1, b1, b2, b3)
                    
            return total % MOD
                
            
            
            
        
        
        total = 0
        options = [(1, 2, 1), (0, 1, 0), (0, 2, 0), (1, 0, 1), (2, 0, 2), (2, 1, 2), (0, 1, 2), (2, 1, 0), (1, 0, 2), (2, 0, 1),
                  (0, 2, 1), (1, 2, 0)]
        
        for a, b, c in options:
            total += calc(1, a, b, c) % MOD
        
        return total % MOD
44100