<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) {
        vector<long long int> result(n+3);
        result[1] = 1;
        result[2] = 2;
        result[3] = 5;
        for(int i=4; i<=n; i++){
            result[i] = (2 * result[i-1] + result[i-3]) % 1000000007;
        }

        return result[n];
    }
};

/*
    n
    
    1 -> 1
    2 -> 2
    3 -> 5
    4 -> 11
    5 -> 24
    6 -> 53
    7 -> 117
    8 -> 258
    9 -> 569
    10 -> 1255

*/
7900