<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 numTilings(int n) {

        unordered_map<int, unsigned int>dp;

        return tilingsUtil(n,n,dp);
    }

    unsigned int tilingsUtil(int firstRow, int secondRow, unordered_map<int, unsigned int>& dp) {
        if(firstRow<0 || secondRow<0) {
            return 0;
        }
        if(firstRow == 0 && secondRow == 0) {
            return 1;
        }

        int hashValue = ((firstRow*10000) + secondRow);

        if(dp.find(hashValue) != dp.end()) {
            return dp[hashValue];
        }

        if(firstRow==secondRow){
            dp[hashValue] = (tilingsUtil(firstRow-2, secondRow, dp) +
                 + tilingsUtil(firstRow-1, secondRow-2, dp) + tilingsUtil(firstRow-2, secondRow-1, dp) + 
                 tilingsUtil(firstRow-1, secondRow-1, dp)) % 1000000007;

            return dp[hashValue];
        }

        if(firstRow==secondRow -1){
            dp[hashValue] = (tilingsUtil(firstRow, secondRow-2, dp) +
                tilingsUtil(firstRow-1, secondRow-2, dp)) % 1000000007;

            return dp[hashValue];
        }

        if(firstRow==secondRow + 1){
            dp[hashValue] = (tilingsUtil(firstRow-2, secondRow, dp) +
                tilingsUtil(firstRow-2, secondRow-1, dp)) % 1000000007;

            return dp[hashValue];
        }

        if(firstRow==secondRow - 2){
            dp[hashValue] = tilingsUtil(firstRow, secondRow-2, dp) % 1000000007;

            return dp[hashValue];
        }

        return 0;

    }
};
13100