<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:
    vector<vector<long long>> memo;
    int numTilings(int n) 
    {
        int mod=1000000007;
        memo=vector<vector<long long>>(n+1,vector<long long>(3,-1));
        return rec(n,0)%mod;    
    }
    int rec(int n,int cur)
    {
        int mod=1000000007;
        if(n==0 && cur==0) return 1;
        if(n<=0) return 0;
        if(memo[n][cur]!=-1) return memo[n][cur];
        long long s1=0;
        long long s2=0;
        long long s3=0;
        if(cur==0)
        {
            s1=rec(n-1,0)%mod;
            s2=rec(n-1,1)%mod;
            s3 = rec(n-1, 2) % mod + rec(n-2, 0) % mod;
        }
        else if(cur==1)
        {
            s1=rec(n-1,2)%mod;
            s2=rec(n-2,0)%mod;
        }
        else{
            s1=rec(n-1,1)%mod;
            s2=rec(n-2,0)%mod;
        }
        return memo[n][cur] = (s1 + s2 + s3) % mod; 
    }
};
9200