<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 = 1_000_000_007
        r = [[-1 for _ in range(3)] for _ in range(n)]
        
        @cache
        def dp(ci, prev):
            
            if ci == n * 3:
                return 1 # reached end
            
            cr = ci // 3 
            cc = ci % 3
            
            taken = set()
            
            if cr > 0 :
                taken.add(prev[cc]) # upr wala
            
            if cc > 0:
                taken.add(prev[cc - 1])
            
            ans = 0
            for i in [0,1,2] : # we will try coloring it will 1 or 2 or 3
                
                if i not in taken:
                    new_prev = list(prev)
                    new_prev[cc] = i
                    ans += dp(ci + 1, tuple(new_prev))
            
            return ans % mod
            
        
        return dp(0, tuple([-1] * 3)) % mod
93300