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Commit 3a72261

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feat: add weekly contest 417 (doocs#3583)
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---
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comments: true
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difficulty: 简单
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3304.Find%20the%20K-th%20Character%20in%20String%20Game%20I/README.md
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---
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<!-- problem:start -->
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# [3304. 找出第 K 个字符 I](https://leetcode.cn/problems/find-the-k-th-character-in-string-game-i)
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[English Version](/solution/3300-3399/3304.Find%20the%20K-th%20Character%20in%20String%20Game%20I/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<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>
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<p>给定一个<strong>正整数</strong> <code>k</code>。</p>
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<p>现在 Bob 会要求 Alice 执行以下操作<strong> 无限次 </strong>:</p>
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<ul>
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<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>
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</ul>
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<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>
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<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>
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<p><strong>注意</strong>,在操作中字符 <code>'z'</code> 可以变成 <code>'a'</code>。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1:</strong></p>
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<div class="example-block">
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<p><strong>输入:</strong><span class="example-io">k = 5</span></p>
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<p><strong>输出:</strong><span class="example-io">"b"</span></p>
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<p><strong>解释:</strong></p>
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<p>最初,<code>word = "a"</code>。需要进行三次操作:</p>
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<ul>
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<li>生成的字符串是 <code>"b"</code>,<code>word</code> 变为 <code>"ab"</code>。</li>
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<li>生成的字符串是 <code>"bc"</code>,<code>word</code> 变为 <code>"abbc"</code>。</li>
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<li>生成的字符串是 <code>"bccd"</code>,<code>word</code> 变为 <code>"abbcbccd"</code>。</li>
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</ul>
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</div>
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<p><strong class="example">示例 2:</strong></p>
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<div class="example-block">
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<p><strong>输入:</strong><span class="example-io">k = 10</span></p>
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<p><strong>输出:</strong><span class="example-io">"c"</span></p>
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</div>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>1 &lt;= k &lt;= 500</code></li>
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</ul>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一:模拟
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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。
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最后返回 $\textit{word}[k - 1]$ 即可。
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时间复杂度 $O(k),ドル空间复杂度 $O(k)$。其中 $k$ 为输入参数。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def kthCharacter(self, k: int) -> str:
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word = [0]
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while len(word) < k:
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word.extend([(x + 1) % 26 for x in word])
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return chr(ord("a") + word[k - 1])
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```
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#### Java
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```java
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class Solution {
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public char kthCharacter(int k) {
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List<Integer> word = new ArrayList<>();
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word.add(0);
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while (word.size() < k) {
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for (int i = 0, m = word.size(); i < m; ++i) {
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word.add((word.get(i) + 1) % 26);
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}
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}
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return (char) ('a' + word.get(k - 1));
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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char kthCharacter(int k) {
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vector<int> word;
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word.push_back(0);
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while (word.size() < k) {
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int m = word.size();
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for (int i = 0; i < m; ++i) {
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word.push_back((word[i] + 1) % 26);
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}
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}
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return 'a' + word[k - 1];
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}
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};
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```
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#### Go
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```go
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func kthCharacter(k int) byte {
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word := []int{0}
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for len(word) < k {
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m := len(word)
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for i := 0; i < m; i++ {
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word = append(word, (word[i]+1)%26)
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}
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}
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return 'a' + byte(word[k-1])
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}
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```
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#### TypeScript
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```ts
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function kthCharacter(k: number): string {
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const word: number[] = [0];
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while (word.length < k) {
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word.push(...word.map(x => (x + 1) % 26));
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}
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return String.fromCharCode(97 + word[k - 1]);
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->
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---
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comments: true
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difficulty: Easy
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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
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---
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<!-- problem:start -->
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# [3304. Find the K-th Character in String Game I](https://leetcode.com/problems/find-the-k-th-character-in-string-game-i)
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[中文文档](/solution/3300-3399/3304.Find%20the%20K-th%20Character%20in%20String%20Game%20I/README.md)
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## Description
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<!-- description:start -->
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<p>Alice and Bob are playing a game. Initially, Alice has a string <code>word = &quot;a&quot;</code>.</p>
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<p>You are given a <strong>positive</strong> integer <code>k</code>.</p>
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<p>Now Bob will ask Alice to perform the following operation <strong>forever</strong>:</p>
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<ul>
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<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>
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</ul>
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<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>
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<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>
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<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>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">k = 5</span></p>
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<p><strong>Output:</strong> <span class="example-io">&quot;b&quot;</span></p>
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<p><strong>Explanation:</strong></p>
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<p>Initially, <code>word = &quot;a&quot;</code>. We need to do the operation three times:</p>
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<ul>
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<li>Generated string is <code>&quot;b&quot;</code>, <code>word</code> becomes <code>&quot;ab&quot;</code>.</li>
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<li>Generated string is <code>&quot;bc&quot;</code>, <code>word</code> becomes <code>&quot;abbc&quot;</code>.</li>
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<li>Generated string is <code>&quot;bccd&quot;</code>, <code>word</code> becomes <code>&quot;abbcbccd&quot;</code>.</li>
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</ul>
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</div>
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<p><strong class="example">Example 2:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">k = 10</span></p>
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<p><strong>Output:</strong> <span class="example-io">&quot;c&quot;</span></p>
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</div>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= k &lt;= 500</code></li>
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</ul>
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<!-- description:end -->
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## Solutions
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<!-- solution:start -->
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### Solution 1: Simulation
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We can use an array $\textit{word}$ to store the string after each operation. When the length of $\textit{word}$ is less than $k,ドル we continuously perform operations on $\textit{word}$.
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Finally, return $\textit{word}[k - 1]$.
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The time complexity is $O(k),ドル and the space complexity is $O(k)$. Here, $k$ is the input parameter.
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def kthCharacter(self, k: int) -> str:
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word = [0]
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while len(word) < k:
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word.extend([(x + 1) % 26 for x in word])
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return chr(ord("a") + word[k - 1])
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```
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#### Java
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```java
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class Solution {
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public char kthCharacter(int k) {
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List<Integer> word = new ArrayList<>();
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word.add(0);
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while (word.size() < k) {
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for (int i = 0, m = word.size(); i < m; ++i) {
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word.add((word.get(i) + 1) % 26);
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}
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}
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return (char) ('a' + word.get(k - 1));
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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char kthCharacter(int k) {
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vector<int> word;
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word.push_back(0);
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while (word.size() < k) {
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int m = word.size();
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for (int i = 0; i < m; ++i) {
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word.push_back((word[i] + 1) % 26);
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}
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}
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return 'a' + word[k - 1];
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}
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};
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```
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#### Go
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```go
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func kthCharacter(k int) byte {
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word := []int{0}
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for len(word) < k {
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m := len(word)
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for i := 0; i < m; i++ {
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word = append(word, (word[i]+1)%26)
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}
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}
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return 'a' + byte(word[k-1])
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}
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```
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#### TypeScript
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```ts
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function kthCharacter(k: number): string {
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const word: number[] = [0];
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while (word.length < k) {
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word.push(...word.map(x => (x + 1) % 26));
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}
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return String.fromCharCode(97 + word[k - 1]);
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->
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class Solution {
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public:
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char kthCharacter(int k) {
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vector<int> word;
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word.push_back(0);
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while (word.size() < k) {
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int m = word.size();
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for (int i = 0; i < m; ++i) {
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word.push_back((word[i] + 1) % 26);
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}
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}
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return 'a' + word[k - 1];
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}
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};
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func kthCharacter(k int) byte {
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word := []int{0}
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for len(word) < k {
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m := len(word)
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for i := 0; i < m; i++ {
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word = append(word, (word[i]+1)%26)
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}
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}
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return 'a' + byte(word[k-1])
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}
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class Solution {
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public char kthCharacter(int k) {
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List<Integer> word = new ArrayList<>();
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word.add(0);
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while (word.size() < k) {
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for (int i = 0, m = word.size(); i < m; ++i) {
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word.add((word.get(i) + 1) % 26);
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}
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}
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return (char) ('a' + word.get(k - 1));
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}
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}
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class Solution:
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def kthCharacter(self, k: int) -> str:
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word = [0]
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while len(word) < k:
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word.extend([(x + 1) % 26 for x in word])
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return chr(ord("a") + word[k - 1])
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function kthCharacter(k: number): string {
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const word: number[] = [0];
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while (word.length < k) {
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word.push(...word.map(x => (x + 1) % 26));
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}
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return String.fromCharCode(97 + word[k - 1]);
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}

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