feat: add weekly contest 417 (#3583)

Author: yanglbme

solution/3300-3399/3304.Find the K-th Character in String Game I/README.md

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+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3304.Find%20the%20K-th%20Character%20in%20String%20Game%20I/README.md
+---
+
+<!-- problem:start -->
+
+# [3304. 找出第 K 个字符 I](https://leetcode.cn/problems/find-the-k-th-character-in-string-game-i)
+
+[English Version](/solution/3300-3399/3304.Find%20the%20K-th%20Character%20in%20String%20Game%20I/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>Alice 和 Bob 正在玩一个游戏。最初，Alice 有一个字符串 <code>word = "a"</code>。</p>
+
+<p>给定一个<strong>正整数</strong> <code>k</code>。</p>
+
+<p>现在 Bob 会要求 Alice 执行以下操作<strong> 无限次 </strong>:</p>
+
+<ul>
+	<li>将 <code>word</code> 中的每个字符<strong> 更改 </strong>为英文字母表中的<strong> 下一个 </strong>字符来生成一个新字符串，并将其<strong> 追加 </strong>到原始的 <code>word</code>。</li>
+</ul>
+
+<p>例如，对 <code>"c"</code> 进行操作生成 <code>"cd"</code>，对 <code>"zb"</code> 进行操作生成 <code>"zbac"</code>。</p>
+
+<p>在执行足够多的操作后， <code>word</code> 中 <strong>至少 </strong>存在 <code>k</code> 个字符，此时返回 <code>word</code> 中第 <code>k</code> 个字符的值。</p>
+
+<p><strong>注意</strong>，在操作中字符 <code>'z'</code> 可以变成 <code>'a'</code>。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1:</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">k = 5</span></p>
+
+<p><strong>输出：</strong><span class="example-io">"b"</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>最初，<code>word = "a"</code>。需要进行三次操作:</p>
+
+<ul>
+	<li>生成的字符串是 <code>"b"</code>，<code>word</code> 变为 <code>"ab"</code>。</li>
+	<li>生成的字符串是 <code>"bc"</code>，<code>word</code> 变为 <code>"abbc"</code>。</li>
+	<li>生成的字符串是 <code>"bccd"</code>，<code>word</code> 变为 <code>"abbcbccd"</code>。</li>
+</ul>
+</div>
+
+<p><strong class="example">示例 2:</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">k = 10</span></p>
+
+<p><strong>输出：</strong><span class="example-io">"c"</span></p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= k &lt;= 500</code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：模拟
+
+我们可以使用一个数组 $\textit{word}$ 来存储每次操作后的字符串，当 $\textit{word}$ 的长度小于 $k$ 时，我们不断地对 $\textit{word}$ 进行操作。
+
+最后返回 $\textit{word}[k - 1]$ 即可。
+
+时间复杂度 $O(k)$，空间复杂度 $O(k)$。其中 $k$ 为输入参数。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def kthCharacter(self, k: int) -> str:
+        word = [0]
+        while len(word) < k:
+            word.extend([(x + 1) % 26 for x in word])
+        return chr(ord("a") + word[k - 1])
+```
+
+#### Java
+
+```java
+class Solution {
+    public char kthCharacter(int k) {
+        List<Integer> word = new ArrayList<>();
+        word.add(0);
+        while (word.size() < k) {
+            for (int i = 0, m = word.size(); i < m; ++i) {
+                word.add((word.get(i) + 1) % 26);
+            }
+        }
+        return (char) ('a' + word.get(k - 1));
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    char kthCharacter(int k) {
+        vector<int> word;
+        word.push_back(0);
+        while (word.size() < k) {
+            int m = word.size();
+            for (int i = 0; i < m; ++i) {
+                word.push_back((word[i] + 1) % 26);
+            }
+        }
+        return 'a' + word[k - 1];
+    }
+};
+```
+
+#### Go
+
+```go
+func kthCharacter(k int) byte {
+	word := []int{0}
+	for len(word) < k {
+		m := len(word)
+		for i := 0; i < m; i++ {
+			word = append(word, (word[i]+1)%26)
+		}
+	}
+	return 'a' + byte(word[k-1])
+}
+```
+
+#### TypeScript
+
+```ts
+function kthCharacter(k: number): string {
+    const word: number[] = [0];
+    while (word.length < k) {
+        word.push(...word.map(x => (x + 1) % 26));
+    }
+    return String.fromCharCode(97 + word[k - 1]);
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3300-3399/3304.Find the K-th Character in String Game I/README_EN.md

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+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3304.Find%20the%20K-th%20Character%20in%20String%20Game%20I/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3304. Find the K-th Character in String Game I](https://leetcode.com/problems/find-the-k-th-character-in-string-game-i)
+
+[中文文档](/solution/3300-3399/3304.Find%20the%20K-th%20Character%20in%20String%20Game%20I/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>Alice and Bob are playing a game. Initially, Alice has a string <code>word = &quot;a&quot;</code>.</p>
+
+<p>You are given a <strong>positive</strong> integer <code>k</code>.</p>
+
+<p>Now Bob will ask Alice to perform the following operation <strong>forever</strong>:</p>
+
+<ul>
+	<li>Generate a new string by <strong>changing</strong> each character in <code>word</code> to its <strong>next</strong> character in the English alphabet, and <strong>append</strong> it to the <em>original</em> <code>word</code>.</li>
+</ul>
+
+<p>For example, performing the operation on <code>&quot;c&quot;</code> generates <code>&quot;cd&quot;</code> and performing the operation on <code>&quot;zb&quot;</code> generates <code>&quot;zbac&quot;</code>.</p>
+
+<p>Return the value of the <code>k<sup>th</sup></code> character in <code>word</code>, after enough operations have been done for <code>word</code> to have <strong>at least</strong> <code>k</code> characters.</p>
+
+<p><strong>Note</strong> that the character <code>&#39;z&#39;</code> can be changed to <code>&#39;a&#39;</code> in the operation.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">k = 5</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">&quot;b&quot;</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>Initially, <code>word = &quot;a&quot;</code>. We need to do the operation three times:</p>
+
+<ul>
+	<li>Generated string is <code>&quot;b&quot;</code>, <code>word</code> becomes <code>&quot;ab&quot;</code>.</li>
+	<li>Generated string is <code>&quot;bc&quot;</code>, <code>word</code> becomes <code>&quot;abbc&quot;</code>.</li>
+	<li>Generated string is <code>&quot;bccd&quot;</code>, <code>word</code> becomes <code>&quot;abbcbccd&quot;</code>.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">k = 10</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">&quot;c&quot;</span></p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= k &lt;= 500</code></li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Simulation
+
+We can use an array $\textit{word}$ to store the string after each operation. When the length of $\textit{word}$ is less than $k$, we continuously perform operations on $\textit{word}$.
+
+Finally, return $\textit{word}[k - 1]$.
+
+The time complexity is $O(k)$, and the space complexity is $O(k)$. Here, $k$ is the input parameter.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def kthCharacter(self, k: int) -> str:
+        word = [0]
+        while len(word) < k:
+            word.extend([(x + 1) % 26 for x in word])
+        return chr(ord("a") + word[k - 1])
+```
+
+#### Java
+
+```java
+class Solution {
+    public char kthCharacter(int k) {
+        List<Integer> word = new ArrayList<>();
+        word.add(0);
+        while (word.size() < k) {
+            for (int i = 0, m = word.size(); i < m; ++i) {
+                word.add((word.get(i) + 1) % 26);
+            }
+        }
+        return (char) ('a' + word.get(k - 1));
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    char kthCharacter(int k) {
+        vector<int> word;
+        word.push_back(0);
+        while (word.size() < k) {
+            int m = word.size();
+            for (int i = 0; i < m; ++i) {
+                word.push_back((word[i] + 1) % 26);
+            }
+        }
+        return 'a' + word[k - 1];
+    }
+};
+```
+
+#### Go
+
+```go
+func kthCharacter(k int) byte {
+	word := []int{0}
+	for len(word) < k {
+		m := len(word)
+		for i := 0; i < m; i++ {
+			word = append(word, (word[i]+1)%26)
+		}
+	}
+	return 'a' + byte(word[k-1])
+}
+```
+
+#### TypeScript
+
+```ts
+function kthCharacter(k: number): string {
+    const word: number[] = [0];
+    while (word.length < k) {
+        word.push(...word.map(x => (x + 1) % 26));
+    }
+    return String.fromCharCode(97 + word[k - 1]);
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3300-3399/3304.Find the K-th Character in String Game I/Solution.cpp

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+class Solution {
+public:
+    char kthCharacter(int k) {
+        vector<int> word;
+        word.push_back(0);
+        while (word.size() < k) {
+            int m = word.size();
+            for (int i = 0; i < m; ++i) {
+                word.push_back((word[i] + 1) % 26);
+            }
+        }
+        return 'a' + word[k - 1];
+    }
+};

solution/3300-3399/3304.Find the K-th Character in String Game I/Solution.go

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+func kthCharacter(k int) byte {
+	word := []int{0}
+	for len(word) < k {
+		m := len(word)
+		for i := 0; i < m; i++ {
+			word = append(word, (word[i]+1)%26)
+		}
+	}
+	return 'a' + byte(word[k-1])
+}

solution/3300-3399/3304.Find the K-th Character in String Game I/Solution.java

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+class Solution {
+    public char kthCharacter(int k) {
+        List<Integer> word = new ArrayList<>();
+        word.add(0);
+        while (word.size() < k) {
+            for (int i = 0, m = word.size(); i < m; ++i) {
+                word.add((word.get(i) + 1) % 26);
+            }
+        }
+        return (char) ('a' + word.get(k - 1));
+    }
+}

solution/3300-3399/3304.Find the K-th Character in String Game I/Solution.py

@@ -0,0 +1,6 @@
+class Solution:
+    def kthCharacter(self, k: int) -> str:
+        word = [0]
+        while len(word) < k:
+            word.extend([(x + 1) % 26 for x in word])
+        return chr(ord("a") + word[k - 1])

solution/3300-3399/3304.Find the K-th Character in String Game I/Solution.ts

@@ -0,0 +1,7 @@
+function kthCharacter(k: number): string {
+    const word: number[] = [0];
+    while (word.length < k) {
+        word.push(...word.map(x => (x + 1) % 26));
+    }
+    return String.fromCharCode(97 + word[k - 1]);
+}