<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 {
public:
    int numTilings(int n) {
        int mod = 1e9+ 7;
        vector<long> dp(10001);
        dp[0] = 1;
        dp[1] = 1;
        dp[2] = 2;
        dp[3] = 5;
        if (n <= 3) {
            return dp[n];
        }
        vector<long> top(10001, 0);
        top[2] = 1;
        vector<long> bottom(10001, 0);
        bottom[2] = 1;
        // dp[3] = 2 + 1
        // dp[n] = dp[n-1] + dp[n-2] + dp[n-3] * 2;
        for (int i = 3; i <=n; i++) {
            dp[i] = (dp[i-1] + dp[i-2] + top[i-1] + bottom[i-1]) % mod;
            top[i] = (dp[i-2] + bottom[i-1]) % mod;
            bottom[i] = (dp[i-2] + top[i-1]) % mod;
        }
        return dp[n] % mod;
    }
};
18100