<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 patterns(self, colors: list, ans: list, last: int):
        if len(colors) == 3:
            ans.append(tuple(colors))
            return
        
        for i in range(3):
            if last is None or i != last:
                colors.append(i)
                self.patterns(colors, ans, i)
                colors.pop()
                
    def dfs(self, i: int, target: int, currPattern: tuple, memo: dict, basePatternToCompatible: dict) -> int:
        if i == target:
            return 1
        
        memoKey = f"{i}-{currPattern}"
        if memoKey in memo:
            return memo[memoKey]
        
        nextPatterns = basePatternToCompatible[currPattern]
        ans = 0
        for pattern in nextPatterns:
            ans += self.dfs(i+1, target, pattern, memo, basePatternToCompatible)
            
        memo[memoKey] = ans
        return memo[memoKey]
    
    def numOfWays(self, n: int) -> int:
        basePatterns = []
        self.patterns([], basePatterns, None)  
        
        basePatternToCompatible = dict()
        for pattern in basePatterns:
            compatiblePatterns = []
            for nextPattern in basePatterns:
                compatible = True
                for i in range(3):
                    if nextPattern[i] == pattern[i]:
                        compatible = False
                        break
                
                if compatible:
                    compatiblePatterns.append(nextPattern)
            basePatternToCompatible[pattern] = compatiblePatterns
            
        ans = 0
        memo = dict()
        for pattern in basePatterns:
            ans += self.dfs(1, n, pattern, memo, basePatternToCompatible)
            
        
        return ans % (10**9 + 7)
        
67300