Sync LeetCode submission Runtime - 6 ms (23.38%), Memory - 19.6 MB (9.77%)

Author: jimit105

0806-domino-and-tromino-tiling/README.md

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+<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>

0806-domino-and-tromino-tiling/solution.py

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+# Approach 1: Dynamic Programming (Top-down)
+
+# Time: O(n)
+# Space: O(n)
+
+class Solution:
+    def numTilings(self, n: int) -> int:
+        MOD = 10 ** 9 + 7
+
+        @cache
+        def p(n):
+            if n == 2:
+                return 1
+            return (p(n - 1) + f(n - 2)) % MOD
+
+        @cache
+        def f(n):
+            if n <= 2:
+                return n
+            return (f(n - 1) + f(n - 2) + 2 * p(n - 1)) % MOD
+
+        return f(n)
+