<p>You have two types of tiles: a <code>2 x 1</code> domino shape and a tromino shape. You may rotate these shapes.</p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/15/lc-domino.jpg" style="width: 362px; height: 195px;" />
<p>Given an integer n, return <em>the number of ways to tile an</em> <code>2 x n</code> <em>board</em>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>In a tiling, every square must be covered by a tile. Two tilings are different if and only if there are two 4-directionally adjacent cells on the board such that exactly one of the tilings has both squares occupied by a tile.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/15/lc-domino1.jpg" style="width: 500px; height: 226px;" />
<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> 5
<strong>Explanation:</strong> The five different ways are show above.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
</ul>

class Solution:

    def numTilings(self, n: int) -> int:
        @cache
        def dfs(i,j):
            # termination condition 
            if i > n or j > n: 
                return 0 
            if i == n and j == n: 
                return 1 

            res = 0 
            # recursion cases:
            if i == j: 
                res = \
                (dfs(i+2, j+2) # placing 2 hori dom \
                + dfs(i+1, j+1) # placing 1 verti dom \ 
                + dfs(i+1, j+2) # placing 1 L shaped trom\
                + dfs(i+2, j+1)) # placing inverted L shaped trom\
                
            elif i > j:
                res = \
                (dfs(i, j+2) # placing 1 hori dom
                + dfs(i+1, j+2)) # placing 1 L trom
                
            else:
                res = \
                (dfs(i+2, j) # placing 1 hori dom\
                + dfs(i+2, j+1)) # placing inverted L shaped trom\
            return res % mod
    
        mod = 10**9 + 7
        return dfs(0, 0)
19200