<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:
    unordered_map<int, long> f1;
    unordered_map<int, long> f2;
    int MOD = 1e9 + 7;
    long p(int n){
        if(f2.find(n) != f2.end())
            return f2[n];
        if(n==2)
            return 1;
        else{
            return f2[n] = (p(n-1) + f(n-2)) % MOD;
            
        }
        
    }
    long f(int n){
        if(f1.find(n) != f1.end())
            return f1[n];
        if(n == 1)
            return 1;
        else if(n==2)
            return 2;
        else{
            return f1[n] = (f(n-1) + f(n-2) + 2 * p(n-1)) % MOD;
        }
        
    }
    int numTilings(int n) {
        return f(n);
    }
};
11200