feat: add new lc problems

solution/3300-3399/3330.Find the Original Typed String I/README.md

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+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3330.Find%20the%20Original%20Typed%20String%20I/README.md
+---
+
+<!-- problem:start -->
+
+# [3330. 找到初始输入字符串 I](https://leetcode.cn/problems/find-the-original-typed-string-i)
+
+[English Version](/solution/3300-3399/3330.Find%20the%20Original%20Typed%20String%20I/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<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>
+
+<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>
+
+<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>
+
+<p>请你返回 Alice 一开始可能想要输入字符串的总方案数。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>word = "abbcccc"</span></p>
+
+<p><span class="example-io"><b>输出：</b>5</span></p>
+
+<p><strong>解释：</strong></p>
+
+<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>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>word = "abcd"</span></p>
+
+<p><span class="example-io"><b>输出：</b>1</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>唯一可能的字符串是&nbsp;<code>"abcd"</code>&nbsp;。</p>
+</div>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>word = "aaaa"</span></p>
+
+<p><span class="example-io"><b>输出：</b>4</span></p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= word.length &lt;= 100</code></li>
+	<li><code>word</code>&nbsp;只包含小写英文字母。</li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def possibleStringCount(self, word: str) -> int:
+        return 1 + sum(x == y for x, y in pairwise(word))
+```
+
+#### Java
+
+```java
+class Solution {
+    public int possibleStringCount(String word) {
+        int f = 1;
+        for (int i = 1; i < word.length(); ++i) {
+            if (word.charAt(i) == word.charAt(i - 1)) {
+                ++f;
+            }
+        }
+        return f;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int possibleStringCount(string word) {
+        int f = 1;
+        for (int i = 1; i < word.size(); ++i) {
+            f += word[i] == word[i - 1];
+        }
+        return f;
+    }
+};
+```
+
+#### Go
+
+```go
+func possibleStringCount(word string) int {
+	f := 1
+	for i := 1; i < len(word); i++ {
+		if word[i] == word[i-1] {
+			f++
+		}
+	}
+	return f
+}
+```
+
+#### TypeScript
+
+```ts
+function possibleStringCount(word: string): number {
+    let f = 1;
+    for (let i = 1; i < word.length; ++i) {
+        f += word[i] === word[i - 1] ? 1 : 0;
+    }
+    return f;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3300-3399/3330.Find the Original Typed String I/README_EN.md

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+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3330.Find%20the%20Original%20Typed%20String%20I/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3330. Find the Original Typed String I](https://leetcode.com/problems/find-the-original-typed-string-i)
+
+[中文文档](/solution/3300-3399/3330.Find%20the%20Original%20Typed%20String%20I/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<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>
+
+<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>
+
+<p>You are given a string <code>word</code>, which represents the <strong>final</strong> output displayed on Alice&#39;s screen.</p>
+
+<p>Return the total number of <em>possible</em> original strings that Alice <em>might</em> have intended to type.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">word = &quot;abbcccc&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">5</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<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>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">word = &quot;abcd&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The only possible string is <code>&quot;abcd&quot;</code>.</p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">word = &quot;aaaa&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">4</span></p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= word.length &lt;= 100</code></li>
+	<li><code>word</code> consists only of lowercase English letters.</li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def possibleStringCount(self, word: str) -> int:
+        return 1 + sum(x == y for x, y in pairwise(word))
+```
+
+#### Java
+
+```java
+class Solution {
+    public int possibleStringCount(String word) {
+        int f = 1;
+        for (int i = 1; i < word.length(); ++i) {
+            if (word.charAt(i) == word.charAt(i - 1)) {
+                ++f;
+            }
+        }
+        return f;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int possibleStringCount(string word) {
+        int f = 1;
+        for (int i = 1; i < word.size(); ++i) {
+            f += word[i] == word[i - 1];
+        }
+        return f;
+    }
+};
+```
+
+#### Go
+
+```go
+func possibleStringCount(word string) int {
+	f := 1
+	for i := 1; i < len(word); i++ {
+		if word[i] == word[i-1] {
+			f++
+		}
+	}
+	return f
+}
+```
+
+#### TypeScript
+
+```ts
+function possibleStringCount(word: string): number {
+    let f = 1;
+    for (let i = 1; i < word.length; ++i) {
+        f += word[i] === word[i - 1] ? 1 : 0;
+    }
+    return f;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3300-3399/3330.Find the Original Typed String I/Solution.cpp

@@ -0,0 +1,10 @@
+class Solution {
+public:
+    int possibleStringCount(string word) {
+        int f = 1;
+        for (int i = 1; i < word.size(); ++i) {
+            f += word[i] == word[i - 1];
+        }
+        return f;
+    }
+};

solution/3300-3399/3330.Find the Original Typed String I/Solution.go

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+func possibleStringCount(word string) int {
+	f := 1
+	for i := 1; i < len(word); i++ {
+		if word[i] == word[i-1] {
+			f++
+		}
+	}
+	return f
+}

solution/3300-3399/3330.Find the Original Typed String I/Solution.java

@@ -0,0 +1,11 @@
+class Solution {
+    public int possibleStringCount(String word) {
+        int f = 1;
+        for (int i = 1; i < word.length(); ++i) {
+            if (word.charAt(i) == word.charAt(i - 1)) {
+                ++f;
+            }
+        }
+        return f;
+    }
+}

solution/3300-3399/3330.Find the Original Typed String I/Solution.py

@@ -0,0 +1,3 @@
+class Solution:
+    def possibleStringCount(self, word: str) -> int:
+        return 1 + sum(x == y for x, y in pairwise(word))

solution/3300-3399/3330.Find the Original Typed String I/Solution.ts

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+function possibleStringCount(word: string): number {
+    let f = 1;
+    for (let i = 1; i < word.length; ++i) {
+        f += word[i] === word[i - 1] ? 1 : 0;
+    }
+    return f;
+}