<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:
    long long dp[1001][1001];
    long long MOD = 1000000007l;
    long long countRecursive(int c1, int c2, int n) {
        if(c1 == n && c2 == n) {
            return 1;
        }
        else if(c1 > n || c2 > n) return 0;
        if(dp[c1][c2] != -1) return dp[c1][c2];
        // vertical domino
        long long cnt = 0ll;
        if(c1 == c2) {
            cnt = countRecursive(c1+1, c2+1, n) % MOD;
        }      
        //horizontal domino cases
        cnt = (cnt + countRecursive(c1+2, c2+2, n)) % MOD;

        //tromino cases
        if(c1 == c2) {
            cnt = (cnt + countRecursive(c1+1, c2+2, n)) % MOD;
            cnt = (cnt + countRecursive(c1+2, c2+1, n)) % MOD;
            // tromino + domino
            cnt = (cnt + countRecursive(c1+3, c2+2, n)) % MOD;
            cnt = (cnt + countRecursive(c1+2, c2+3, n)) % MOD;
        }
        if(c1 == c2 + 1) cnt = (cnt + countRecursive(c1+1, c2+2, n)) % MOD;
        if(c1 == c2 - 1) cnt = (cnt + countRecursive(c1+2, c2+1, n)) % MOD;
        dp[c1][c2] = cnt% MOD;
        return dp[c1][c2];
    }
    int numTilings(int n) {
        memset(dp, -1, sizeof(dp));
        return countRecursive(0, 0, n);
    }
};
16400