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Commit 7fc3166

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feat: add new lc problems (doocs#3676)
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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/3300-3399/3330.Find%20the%20Original%20Typed%20String%20I/README.md
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---
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<!-- problem:start -->
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# [3330. 找到初始输入字符串 I](https://leetcode.cn/problems/find-the-original-typed-string-i)
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[English Version](/solution/3300-3399/3330.Find%20the%20Original%20Typed%20String%20I/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>Alice 正在她的电脑上输入一个字符串。但是她打字技术比较笨拙,她&nbsp;<strong>可能</strong>&nbsp;在一个按键上按太久,导致一个字符被输入&nbsp;<strong>多次</strong>&nbsp;。</p>
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<p>尽管 Alice 尽可能集中注意力,她仍然可能会犯错 <strong>至多</strong>&nbsp;一次。</p>
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<p>给你一个字符串&nbsp;<code>word</code> ,它表示 <strong>最终</strong>&nbsp;显示在 Alice 显示屏上的结果。</p>
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<p>请你返回 Alice 一开始可能想要输入字符串的总方案数。</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><span class="example-io"><b>输入:</b>word = "abbcccc"</span></p>
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<p><span class="example-io"><b>输出:</b>5</span></p>
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<p><strong>解释:</strong></p>
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<p>可能的字符串包括:<code>"abbcccc"</code>&nbsp;,<code>"abbccc"</code>&nbsp;,<code>"abbcc"</code>&nbsp;,<code>"abbc"</code>&nbsp;和&nbsp;<code>"abcccc"</code>&nbsp;。</p>
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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><span class="example-io"><b>输入:</b>word = "abcd"</span></p>
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<p><span class="example-io"><b>输出:</b>1</span></p>
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<p><strong>解释:</strong></p>
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<p>唯一可能的字符串是&nbsp;<code>"abcd"</code>&nbsp;。</p>
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</div>
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<p><strong class="example">示例 3:</strong></p>
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<div class="example-block">
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<p><span class="example-io"><b>输入:</b>word = "aaaa"</span></p>
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<p><span class="example-io"><b>输出:</b>4</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;= word.length &lt;= 100</code></li>
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<li><code>word</code>&nbsp;只包含小写英文字母。</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
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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 possibleStringCount(self, word: str) -> int:
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return 1 + sum(x == y for x, y in pairwise(word))
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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 int possibleStringCount(String word) {
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int f = 1;
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for (int i = 1; i < word.length(); ++i) {
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if (word.charAt(i) == word.charAt(i - 1)) {
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++f;
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}
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}
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return f;
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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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int possibleStringCount(string word) {
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int f = 1;
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for (int i = 1; i < word.size(); ++i) {
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f += word[i] == word[i - 1];
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}
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return f;
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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 possibleStringCount(word string) int {
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f := 1
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for i := 1; i < len(word); i++ {
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if word[i] == word[i-1] {
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f++
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}
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}
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return f
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}
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```
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#### TypeScript
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```ts
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function possibleStringCount(word: string): number {
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let f = 1;
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for (let i = 1; i < word.length; ++i) {
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f += word[i] === word[i - 1] ? 1 : 0;
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}
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return f;
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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/3300-3399/3330.Find%20the%20Original%20Typed%20String%20I/README_EN.md
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---
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<!-- problem:start -->
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# [3330. Find the Original Typed String I](https://leetcode.com/problems/find-the-original-typed-string-i)
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[中文文档](/solution/3300-3399/3330.Find%20the%20Original%20Typed%20String%20I/README.md)
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## Description
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<!-- description:start -->
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<p>Alice is attempting to type a specific string on her computer. However, she tends to be clumsy and <strong>may</strong> press a key for too long, resulting in a character being typed <strong>multiple</strong> times.</p>
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<p>Although Alice tried to focus on her typing, she is aware that she may still have done this <strong>at most</strong> <em>once</em>.</p>
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<p>You are given a string <code>word</code>, which represents the <strong>final</strong> output displayed on Alice&#39;s screen.</p>
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<p>Return the total number of <em>possible</em> original strings that Alice <em>might</em> have intended to type.</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">word = &quot;abbcccc&quot;</span></p>
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<p><strong>Output:</strong> <span class="example-io">5</span></p>
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<p><strong>Explanation:</strong></p>
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<p>The possible strings are: <code>&quot;abbcccc&quot;</code>, <code>&quot;abbccc&quot;</code>, <code>&quot;abbcc&quot;</code>, <code>&quot;abbc&quot;</code>, and <code>&quot;abcccc&quot;</code>.</p>
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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">word = &quot;abcd&quot;</span></p>
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<p><strong>Output:</strong> <span class="example-io">1</span></p>
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<p><strong>Explanation:</strong></p>
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<p>The only possible string is <code>&quot;abcd&quot;</code>.</p>
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</div>
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<p><strong class="example">Example 3:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">word = &quot;aaaa&quot;</span></p>
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<p><strong>Output:</strong> <span class="example-io">4</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;= word.length &lt;= 100</code></li>
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<li><code>word</code> consists only of lowercase English letters.</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
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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 possibleStringCount(self, word: str) -> int:
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return 1 + sum(x == y for x, y in pairwise(word))
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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 int possibleStringCount(String word) {
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int f = 1;
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for (int i = 1; i < word.length(); ++i) {
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if (word.charAt(i) == word.charAt(i - 1)) {
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++f;
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}
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}
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return f;
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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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int possibleStringCount(string word) {
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int f = 1;
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for (int i = 1; i < word.size(); ++i) {
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f += word[i] == word[i - 1];
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}
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return f;
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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 possibleStringCount(word string) int {
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f := 1
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for i := 1; i < len(word); i++ {
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if word[i] == word[i-1] {
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f++
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}
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}
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return f
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}
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```
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#### TypeScript
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```ts
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function possibleStringCount(word: string): number {
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let f = 1;
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for (let i = 1; i < word.length; ++i) {
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f += word[i] === word[i - 1] ? 1 : 0;
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}
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return f;
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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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int possibleStringCount(string word) {
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int f = 1;
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for (int i = 1; i < word.size(); ++i) {
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f += word[i] == word[i - 1];
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}
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return f;
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}
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};
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func possibleStringCount(word string) int {
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f := 1
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for i := 1; i < len(word); i++ {
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if word[i] == word[i-1] {
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f++
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}
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}
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return f
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}
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class Solution {
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public int possibleStringCount(String word) {
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int f = 1;
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for (int i = 1; i < word.length(); ++i) {
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if (word.charAt(i) == word.charAt(i - 1)) {
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++f;
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}
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}
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return f;
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}
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}
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class Solution:
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def possibleStringCount(self, word: str) -> int:
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return 1 + sum(x == y for x, y in pairwise(word))
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function possibleStringCount(word: string): number {
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let f = 1;
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for (let i = 1; i < word.length; ++i) {
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f += word[i] === word[i - 1] ? 1 : 0;
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}
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return f;
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}

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