<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;

    long long dfs(int x, int y, map<pair<int,int>, int> &dp){
        if (x == 0 && y == 0) return 1;
        if (x < 0 || y < 0) return 0;
        if (dp.find({x, y}) != dp.end()) return dp[{x, y}];
        long long ans = 0;
        if (x != y){
            if (x < y){
                ans += dfs(x, y - 2, dp);
            }
            else {
                ans += dfs(x - 2, y, dp);
            }
        }
        else {
            ans += dfs(x - 2, y - 2, dp);
            ans += dfs(x - 1, y - 1, dp);
            ans += dfs(x - 1, y - 2, dp);
            ans += dfs(x - 2, y - 1, dp);
        }
        if (y - 1 == x - 2){
            ans += dfs(x - 2, y - 1, dp);
        }
        if (y - 2 == x - 1){
            ans += dfs(x - 1, y - 2, dp);
        }
        return dp[{x, y}] = ans % mod;
    }

    int numTilings(int n) {
        map<pair<int,int>, int> dp;
        return dfs(n, n, dp);
    }
};
13600