<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:
    #define ll long long
    const ll MOD = 1e9+7;
    int numTilings(int n) {
        if (n == 1) return 1;
        if (n == 2) return 2;
        vector<ll> dp(n+1, 0);
        dp[1] = 1, dp[2] = 2, dp[3] = 5;
        for (int i=4; i<=n; i++)
            dp[i] = ((2 * dp[i-1]) % MOD + dp[i-3]) % MOD;
        return dp[n];
    }                           // TC = O(n) SC = O(n)
};
/*
f(n) = f(n-1) + f(n-2) + 2 * f(n-3) + 2 * f(n-4) + ...... + 2 * f(0)
Substituting n = n-1
f(n-1) = f(n-2) + f(n-3) + 2 * f(n-4) + 2 * f (n-5) + ... + 2 * f(0)
Subtracting both the equations:
=> f(n) - f(n-1) = f(n-1) + f(n-3)
=> f(n) = 2 * f(n-1) + f(n-3), n > 5
For n <= 5, one can find it manually.
*/
8000