<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 getNext(self, current_tuple):
        unique_next = [(1,2,1),(1,2,0),(1,0,1), (2,0,1)]
        duplicate_next = [ (1,0,1), (2,0,1), (1,0,2), (2,0,2), (1,2,1)]
        def rotate(current_tuple, duplicate = False):
            result = []
            translate_map = []
            if duplicate == False:
                translate_map = list(current_tuple)
                for tup in unique_next:
                    result.append((translate_map[tup[0]], translate_map[tup[1]], translate_map[tup[2]]))
            else:
                translate_map = [current_tuple[0], current_tuple[1], (3 - current_tuple[0] - current_tuple[1])]
                for tup in duplicate_next:
                    result.append((translate_map[tup[0]], translate_map[tup[1]], translate_map[tup[2]]))
            
            return result
                
        if current_tuple == (0, 1, 2):
            return unique_next
        elif current_tuple == (0,1,0):
            return duplicate_next
        if current_tuple[0] == current_tuple[2]:
            return rotate(current_tuple, duplicate = True)
        else:
            return rotate(current_tuple)
    
            
    def numOfWays(self, n: int) -> int:
        if n == 1:
            return 12
        cached_result = {(0,1,0):1, (1,0,1):1, (2,0,1):1,(0,1,2):1, (1,0,2):1, (2,0,2):1, (0,2,0):1, (1,2,0):1, (2,1,0):1, (0,2,1):1, (1,2,1):1, (2,1,2):1}
        
        key_to_next = {}
        for key in cached_result:
            next_for_key = self.getNext(key)
            key_to_next[key] = next_for_key
        # print(f"key_to_next {key_to_next}")
        
        for i in range(1, n):
            new_result = {(0,1,0):0, (1,0,1):0, (2,0,1):0,(0,1,2):0, (1,0,2):0, (2,0,2):0, (0,2,0):0, (1,2,0):0, (2,1,0):0, (0,2,1):0, (1,2,1):0, (2,1,2):0}
            for key in cached_result:
                next_for_key = key_to_next[key]
                for next_item in next_for_key:
                    new_result[next_item] += cached_result[key]
            #print(f"new_result is {new_result}")
            cached_result = new_result
            #print(f"{i}: {sum(cached_result.values())}")
        
        return sum(cached_result.values()) % (10**9 + 7)
        
            
                    
                    
        
        
        
17100