<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>

MOD = (10 ** 9) + 7

class Solution:
    def numOfWays(self, n: int) -> int:
        
        # Using integers to represent colors
        colors = [1, 2, 3]

        # Memoization dictionary
        memo = {}

        # Helper function to check if two rows are valid together
        def is_valid_row(prev_row, curr_row):
            for i in range(3):
                if prev_row[i] == curr_row[i]:
                    return False  # Adjacent cells can't have the same color
            return True

        # Recursive function to count the valid ways
        def rec_count(row, prev_row):
            if row == n:  # If we've completed all rows, return 1 (valid grid)
                return 1

            # Memoization lookup
            if (row, tuple(prev_row)) in memo:
                return memo[(row, tuple(prev_row))]

            count = 0

            # Try all possible colorings for the current row
            c = set(colors)
            for color1 in c:
                for color2 in c - set([color1]):
                    for color3 in c - set([color2]):
                        
                        curr_row = [color1, color2, color3]
                        
                        if is_valid_row(prev_row, curr_row):
                            count += rec_count(row + 1, curr_row)
                            count %= MOD

            # Memoize the result
            memo[(row, tuple(prev_row))] = count % MOD
            return count
        
        # Start with all possible rows for the first row (the previous row is empty)
        return rec_count(0, [-1, -1, -1])  # Start with an invalid "previous row"

        
38400