<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 int MOD = 1e9 + 7; // Modulo value

    // Memoization table to store already computed states
    vector<vector<vector<int>>> dp;

    int solve(int i, int up, int down, int n) {
        if (i >= n) {
            // Base case: if we've reached the end
            if (i == n && up == 1 && down == 1)
                return 1; // Fully filled
            return 0;     // Otherwise invalid
        }

        // Check if this state is already computed
        if (dp[i][up][down] != -1)
            return dp[i][up][down];

        int ans = 0;

        if (up == 1 && down == 1) {
            // Option 1: Place vertical domino
            ans = (ans + solve(i + 1, 1, 1, n)) % MOD;
            // Option 2: Place horizontal domino on top row
            ans = (ans + solve(i + 1, 1, 0, n)) % MOD;
            // Option 3: Place horizontal domino on bottom row
            ans = (ans + solve(i + 1, 0, 1, n)) % MOD;
            // Option 4: Place two vertical dominoes (or trominoes)
            ans = (ans + solve(i + 2, 1, 1, n)) % MOD;
        }

        if (up == 1 && down == 0) {
            // Place a horizontal domino on the bottom to finish the row
            ans = (ans + solve(i + 1, 0, 1, n)) % MOD;
            // Place a tromino
            ans = (ans + solve(i + 2, 1, 1, n)) % MOD;
        }

        if (up == 0 && down == 1) {
            // Place a horizontal domino on the top to finish the row
            ans = (ans + solve(i + 1, 1, 0, n)) % MOD;
            // Place a tromino
            ans = (ans + solve(i + 2, 1, 1, n)) % MOD;
        }

        // Store the computed answer in the memoization table
        return dp[i][up][down] = ans;
    }

    int numTilings(int n) {
        // Initialize memoization table with -1 (indicating uncomputed states)
        dp = vector<vector<vector<int>>>(
            n + 1, vector<vector<int>>(2, vector<int>(2, -1)));
        // Start solving from the first column (i = 0) with both rows unfilled
        return solve(0, 1, 1, n);
    }
};

10600