<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 dp[1001][1001];
    int modu = 1e9+7;

    int fn(int i, int j, int n){
        if(i==n && j==n){
            return 1;
        }
        if(i>=n || j>=n){
            return 0;
        }
        if(dp[i][j]!=-1){
            return dp[i][j]%modu;
        }

        if(i==j){
            
            int a = fn(i+2,j+1,n)%modu;
            int b = fn(i+1,j+2,n)%modu;
            int c = fn(i+1,j+1,n)%modu;
            int d = fn(i+2,j+2,n)%modu;
            return dp[i][j] = ((a+b)%modu + (c+d)%modu)%modu;
        }
        else if(i==j-1){
            return dp[i][j]=((fn(i+2,j+1,n)%modu) + (fn(i+2,j,n)%modu))%modu;
        }
        else if(j==i-1){
            return dp[i][j]=(fn(i+1,j+2,n)%modu + fn(i,j+2,n)%modu)%modu;
        }
        return dp[i][j]=0;
    }

    int numTilings(int n) {
        // int dp[n+1];
        // memset(dp,0,sizeof(dp));
        // for(int i=1;i<=n;i++){
        //     if(i==1){
        //         dp[i] = 1;
        //     }
        //     else if(i==2){
        //         dp[i] = 2;
        //     }
        //     else{
        //         dp[i] = dp[i-1]+1;
        //     }
        // }

        // return dp[n];
        memset(dp, -1, sizeof(dp));
        return fn(0,0,n);
    }
};
11800