<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<int>> dp;
    int mod = 1e9+7;
    int x(int n,int t)
    {
        if(n == 1) return 1;

        if(n == 2 && t == 1) return 1;
        if(n == 2) return 2;
        if(dp[n][t] != -1) return dp[n][t];
        long long a = 0;
        if(t == 0)
        {
            a = ((2*x(n-1,1)%mod + x(n-2,0)%mod)%mod+x(n-1,0)%mod)%mod;
        }else{
            a = x(n-1,1)%mod + x(n-2,0)%mod;
        }
        return dp[n][t] = a%mod;
    }
    int numTilings(int n) {
        dp.resize(1001,vector<int>(2,-1));
       return x(n,0);
    }
};
10400