Merge pull request #14 from iamAntimPal/Leetcode-75

Leetcode 75

Author: iamAntimPal

Solution/790. Domino and Tromino Tiling/790. Domino and Tromino Tiling.py

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+class Solution:
+    def numTilings(self, n: int) -> int:
+        f = [1, 0, 0, 0]
+        mod = 10**9 + 7
+        for i in range(1, n + 1):
+            g = [0] * 4
+            g[0] = (f[0] + f[1] + f[2] + f[3]) % mod
+            g[1] = (f[2] + f[3]) % mod
+            g[2] = (f[1] + f[3]) % mod
+            g[3] = f[0]
+            f = g
+        return f[0]

Solution/790. Domino and Tromino Tiling/readme.md

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+<!-- problem:start -->
+
+# [790. Domino and Tromino Tiling](https://leetcode.com/problems/domino-and-tromino-tiling)
+
+---
+- **comments**: true
+- **difficulty**: Medium
+- **tags**:
+    - Dynamic Programming
+---
+
+## Description
+
+<!-- description:start -->
+
+<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0790.Domino%20and%20Tromino%20Tiling/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0790.Domino%20and%20Tromino%20Tiling/images/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>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Dynamic Programming
+
+First, we need to understand the problem. The problem is essentially asking us to find the number of ways to tile a $2 \times n$ board, where each square on the board can only be covered by one tile.
+
+There are two types of tiles: `2 x 1` and `L` shapes, and both types of tiles can be rotated. We denote the rotated tiles as `1 x 2` and `L'` shapes.
+
+We define $f[i][j]$ to represent the number of ways to tile the first $2 \times i$ board, where $j$ represents the state of the last column. The last column has 4 states:
+
+-   The last column is fully covered, denoted as $0$
+-   The last column has only the top square covered, denoted as $1$
+-   The last column has only the bottom square covered, denoted as $2$
+-   The last column is not covered, denoted as $3$
+
+The answer is $f[n][0]$. Initially, $f[0][0] = 1$ and the rest $f[0][j] = 0$.
+
+We consider tiling up to the $i$-th column and look at the state transition equations:
+
+When $j = 0$, the last column is fully covered. It can be transitioned from the previous column's states $0, 1, 2, 3$ by placing the corresponding tiles, i.e., $f[i-1][0]$ with a `1 x 2` tile, $f[i-1][1]$ with an `L'` tile, $f[i-1][2]$ with an `L'` tile, or $f[i-1][3]$ with two `2 x 1` tiles. Therefore, $f[i][0] = \sum_{j=0}^3 f[i-1][j]$.
+
+When $j = 1$, the last column has only the top square covered. It can be transitioned from the previous column's states $2, 3$ by placing a `2 x 1` tile or an `L` tile. Therefore, $f[i][1] = f[i-1][2] + f[i-1][3]$.
+
+When $j = 2$, the last column has only the bottom square covered. It can be transitioned from the previous column's states $1, 3$ by placing a `2 x 1` tile or an `L'` tile. Therefore, $f[i][2] = f[i-1][1] + f[i-1][3]$.
+
+When $j = 3$, the last column is not covered. It can be transitioned from the previous column's state $0$. Therefore, $f[i][3] = f[i-1][0]$.
+
+We can see that the state transition equations only involve the previous column's states, so we can use a rolling array to optimize the space complexity.
+
+Note that the values of the states can be very large, so we need to take modulo $10^9 + 7$.
+
+The time complexity is $O(n)$, and the space complexity is $O(1)$. Where $n$ is the number of columns of the board.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def numTilings(self, n: int) -> int:
+        f = [1, 0, 0, 0]
+        mod = 10**9 + 7
+        for i in range(1, n + 1):
+            g = [0] * 4
+            g[0] = (f[0] + f[1] + f[2] + f[3]) % mod
+            g[1] = (f[2] + f[3]) % mod
+            g[2] = (f[1] + f[3]) % mod
+            g[3] = f[0]
+            f = g
+        return f[0]
+```
+
+#### Java
+
+```java
+class Solution {
+    public int numTilings(int n) {
+        long[] f = {1, 0, 0, 0};
+        int mod = (int) 1e9 + 7;
+        for (int i = 1; i <= n; ++i) {
+            long[] g = new long[4];
+            g[0] = (f[0] + f[1] + f[2] + f[3]) % mod;
+            g[1] = (f[2] + f[3]) % mod;
+            g[2] = (f[1] + f[3]) % mod;
+            g[3] = f[0];
+            f = g;
+        }
+        return (int) f[0];
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    const int mod = 1e9 + 7;
+
+    int numTilings(int n) {
+        long f[4] = {1, 0, 0, 0};
+        for (int i = 1; i <= n; ++i) {
+            long g[4] = {0, 0, 0, 0};
+            g[0] = (f[0] + f[1] + f[2] + f[3]) % mod;
+            g[1] = (f[2] + f[3]) % mod;
+            g[2] = (f[1] + f[3]) % mod;
+            g[3] = f[0];
+            memcpy(f, g, sizeof(g));
+        }
+        return f[0];
+    }
+};
+```
+
+#### Go
+
+```go
+func numTilings(n int) int {
+	f := [4]int{}
+	f[0] = 1
+	const mod int = 1e9 + 7
+	for i := 1; i <= n; i++ {
+		g := [4]int{}
+		g[0] = (f[0] + f[1] + f[2] + f[3]) % mod
+		g[1] = (f[2] + f[3]) % mod
+		g[2] = (f[1] + f[3]) % mod
+		g[3] = f[0]
+		f = g
+	}
+	return f[0]
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->