<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 mod = 1e9 + 7;
    typedef long long ll;
    ll dfs(int n, int index, bool tVacant, bool bVacant, vector<vector<vector<int>>> &memo) {
        if (index == n) return !tVacant && !bVacant;
        if (index > n) return 0;
        if (memo[index][tVacant][bVacant] >= 0) return memo[index][tVacant][bVacant] % mod;

        ll total = 0;
        if (tVacant) {
            ll hd = dfs(n, index+1, false, true, memo);
            ll rot = dfs(n, index+2, false, false, memo);
            total = hd + rot;
        } else if (bVacant) {
            ll hd = dfs(n, index+1, true, false, memo);
            ll rut = dfs(n, index+2, false, false, memo);
            total = hd + rut;
        } else {
            ll vd = dfs(n, index+1, false, false, memo);
            ll hd = dfs(n, index+2, false, false, memo);
            
            ll ot = dfs(n, index+1, false, true, memo);
            ll ut = dfs(n, index+1, true, false, memo);
            total = vd + hd + ot + ut;
        }
        memo[index][tVacant][bVacant] = total;
        return total % mod;
    }
    int numTilings(int n) {
        vector<vector<vector<int>>> memo(n, vector<vector<int>>(2, vector<int>(2, -1)));
        return dfs(n, 0, false, false, memo);
    }
};
10500