<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 MOD = 1'000'000'007;

    unordered_map<int, long> f_cache;  
    unordered_map<int, long> p_cache;  
    
    long p(int n) {  
        if (p_cache.find(n) != p_cache.end()) {
            return p_cache[n];
        }
        long val;
        if (n == 2) {
            val = 1L;
        } else {
            val = (p(n - 1) + f(n - 2)) % MOD;
        }
        p_cache[n] = val;
        return val;
    };
    
    long f(int n) {  
        if (f_cache.find(n) != f_cache.end()) {
            return f_cache[n];
        }
        long val;
        if (n == 1) {
            val = 1L;
        } else if (n == 2) {
            val = 2L;
        } else {
            val = (f(n - 1) + f(n - 2) + 2 * p(n - 1)) % MOD;
        }
        f_cache[n] = val;
        return val;
    };
    
    int numTilings(int n) {
        return static_cast<int>(f(n));
    }
};
11400