<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
        dp_sym = [1, 1, 2]
        dp_unsym = [0, 0, 1]
        if n < 3:
            return dp_sym[n]
        for i in range(3, n + 1):
            ways_sym = (
                dp_sym[i - 1]
                + dp_sym[i - 2]
                + 2 * dp_unsym[i - 1]
            )
            dp_sym.append(ways_sym % MOD)
            ways_unsym = (
                dp_sym[i - 2]
                + dp_unsym[i - 1]
            )
            dp_unsym.append(ways_unsym % MOD)
        print(dp_sym)
        print(dp_unsym)
        return dp_sym[-1]
        
16700