<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:
    const long long M = 1e9 + 7;
    unordered_map<int, unordered_map<int, int>> dp;
    int solveDP(int i, int j) {
        if (i < 0 || j < 0)
            return 0;
        if (i == 0 and j == 0)
            return 1;
        if (dp.count(i) and dp[i].count(j)) {
            return dp[i][j];
        }
        long long ans = 0;
        if (i - j == 0) {
            ans = (ans % M + solveDP(i - 1, j - 1) % M +
                   solveDP(i - 1, j - 2) % M + solveDP(i - 2, j - 1) % M +
                   solveDP(i - 2, j - 2) % M) %
                  M;
        } else if (i > j) {
            ans =
                (ans % M + solveDP(i - 2, j) % M + solveDP(i - 2, j - 1) % M) %
                M;
        } else {
            ans =
                (ans % M + solveDP(i, j - 2) % M + solveDP(i - 1, j - 2) % M) %
                M;
        }
        cout << ans << " " << i << " - " << j << endl;
        return dp[i][j] = ans % M;
    }
    int numTilings(int n) { return solveDP(n, n); }
};
15800