<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:
       #(idx,prev1,prev2,prev3)
       # choose all combs of 3 -> fail any that dont word
        def dfs(i,p1,p2,p3):
            if i == n:
                return 1
            key = (i,p1,p2,p3)
            if key in cache:
                return cache[key]
            ways = 0
            for comb in combs:
                if comb[0] == p1 or comb[1] == p2 or comb[2] == p3:
                    continue
                
                ways = (ways + dfs(i+1,comb[0],comb[1],comb[2])) % mod
            
            cache[key] = ways
            return ways
        
        mod = 10**9+7
        cache = {}
        res = 0
        combs = [[1,2,3],[1,2,1],[1,3,2],[1,3,1],[2,1,3],[2,1,2],[2,3,1],[2,3,2],[3,1,2],[3,2,3],[3,2,1],[3,1,3]]
        for p1,p2,p3 in combs:
            res = (res + dfs(1,p1,p2,p3)) % mod
        return res
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