<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>

#define ll long long
class Solution {
public:
    int mod=1e9+7,n;
    ll dp[1002][1002];
    ll rec(int i,int j){
        if(i>n||j>n)return 0;
        if(i==n&&j==n)return 1;
        auto &ans=dp[i][j];
        if(ans!=-1)return ans;
        ans=0;
        if(i==j)ans=(rec(i+2,j+2)+rec(i+1,j+1)+rec(i+1,j+2)+rec(i+2,j+1))%mod;
        else if(i+1==j)ans=(ans+rec(i+2,j+1)+rec(i+2,j))%mod;
        else if(i==j+1)ans=(ans+rec(i+1,j+2)+rec(i,j+2))%mod;
        return ans;
    }
    int numTilings(int N) {
        n=N;
        memset(dp,-1,sizeof(dp));
        return rec(0,0);
    }
};
16300