<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:
        MOD = 10**9 + 7

        def numTilingsMemo(n, memo={}):
            
            if n == 1:
                return 1
            if n == 2:
                return 2
            if n == 3:
                return 5

            if n in memo:
                return memo[n]

            memo[n] =  (2 * numTilingsMemo(n-1) + numTilingsMemo(n-3)) % MOD
            
            return memo[n]

        return numTilingsMemo(n)
        
        # MOD = 10**9 + 7
    
        # if n == 1:
        #     return 1
        # if n == 2:
        #     return 2
        # if n == 3:
        #     return 5

        # dp = [0] * (n + 1)
        # dp[1] = 1
        # dp[2] = 2
        # dp[3] = 5
        
        # for i in range(4, n + 1):
        #     dp[i] = (2 * dp[i-1] + dp[i-3]) % MOD
        
        # return dp[n]
17500