feat: add solutions to lc problems: No.2455~2458

* No.2455.Average Value of Even Numbers That Are Divisible by Three
* No.2456.Most Popular Video Creator
* No.2457.Minimum Addition to Make Integer Beautiful
* No.2458.Height of Binary Tree After Subtree Removal Queries

Author: yanglbme

solution/2400-2499/2451.Odd String Difference/README.md

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+# [2451. 差值数组不同的字符串](https://leetcode.cn/problems/odd-string-difference)
+
+[English Version](/solution/2400-2499/2451.Odd%20String%20Difference/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个字符串数组 <code>words</code>&nbsp;，每一个字符串长度都相同，令所有字符串的长度都为 <code>n</code>&nbsp;。</p>
+
+<p>每个字符串&nbsp;<code>words[i]</code>&nbsp;可以被转化为一个长度为&nbsp;<code>n - 1</code>&nbsp;的&nbsp;<strong>差值整数数组</strong>&nbsp;<code>difference[i]</code>&nbsp;，其中对于&nbsp;<code>0 &lt;= j &lt;= n - 2</code>&nbsp;有&nbsp;<code>difference[i][j] = words[i][j+1] - words[i][j]</code>&nbsp;。注意两个字母的差值定义为它们在字母表中&nbsp;<strong>位置</strong>&nbsp;之差，也就是说&nbsp;<code>'a'</code>&nbsp;的位置是&nbsp;<code>0</code>&nbsp;，<code>'b'</code>&nbsp;的位置是&nbsp;<code>1</code>&nbsp;，<code>'z'</code>&nbsp;的位置是&nbsp;<code>25</code>&nbsp;。</p>
+
+<ul>
+	<li>比方说，字符串&nbsp;<code>"acb"</code>&nbsp;的差值整数数组是&nbsp;<code>[2 - 0, 1 - 2] = [2, -1]</code>&nbsp;。</li>
+</ul>
+
+<p><code>words</code>&nbsp;中所有字符串 <strong>除了一个字符串以外</strong>&nbsp;，其他字符串的差值整数数组都相同。你需要找到那个不同的字符串。</p>
+
+<p>请你返回<em>&nbsp;</em><code>words</code>中&nbsp;<strong>差值整数数组</strong>&nbsp;不同的字符串。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>words = ["adc","wzy","abc"]
+<b>输出：</b>"abc"
+<b>解释：</b>
+- "adc" 的差值整数数组是 [3 - 0, 2 - 3] = [3, -1] 。
+- "wzy" 的差值整数数组是 [25 - 22, 24 - 25]= [3, -1] 。
+- "abc" 的差值整数数组是 [1 - 0, 2 - 1] = [1, 1] 。
+不同的数组是 [1, 1]，所以返回对应的字符串，"abc"。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>words = ["aaa","bob","ccc","ddd"]
+<b>输出：</b>"bob"
+<b>解释：</b>除了 "bob" 的差值整数数组是 [13, -13] 以外，其他字符串的差值整数数组都是 [0, 0] 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>3 &lt;= words.length &lt;= 100</code></li>
+	<li><code>n == words[i].length</code></li>
+	<li><code>2 &lt;= n &lt;= 20</code></li>
+	<li><code>words[i]</code>&nbsp;只含有小写英文字母。</li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2400-2499/2451.Odd String Difference/README_EN.md

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+# [2451. Odd String Difference](https://leetcode.com/problems/odd-string-difference)
+
+[中文文档](/solution/2400-2499/2451.Odd%20String%20Difference/README.md)
+
+## Description
+
+<p>You are given an array of equal-length strings <code>words</code>. Assume that the length of each string is <code>n</code>.</p>
+
+<p>Each string <code>words[i]</code> can be converted into a <strong>difference integer array</strong> <code>difference[i]</code> of length <code>n - 1</code> where <code>difference[i][j] = words[i][j+1] - words[i][j]</code> where <code>0 &lt;= j &lt;= n - 2</code>. Note that the difference between two letters is the difference between their <strong>positions</strong> in the alphabet i.e.&nbsp;the position of <code>&#39;a&#39;</code> is <code>0</code>, <code>&#39;b&#39;</code> is <code>1</code>, and <code>&#39;z&#39;</code> is <code>25</code>.</p>
+
+<ul>
+	<li>For example, for the string <code>&quot;acb&quot;</code>, the difference integer array is <code>[2 - 0, 1 - 2] = [2, -1]</code>.</li>
+</ul>
+
+<p>All the strings in words have the same difference integer array, <strong>except one</strong>. You should find that string.</p>
+
+<p>Return<em> the string in </em><code>words</code><em> that has different <strong>difference integer array</strong>.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;adc&quot;,&quot;wzy&quot;,&quot;abc&quot;]
+<strong>Output:</strong> &quot;abc&quot;
+<strong>Explanation:</strong> 
+- The difference integer array of &quot;adc&quot; is [3 - 0, 2 - 3] = [3, -1].
+- The difference integer array of &quot;wzy&quot; is [25 - 22, 24 - 25]= [3, -1].
+- The difference integer array of &quot;abc&quot; is [1 - 0, 2 - 1] = [1, 1]. 
+The odd array out is [1, 1], so we return the corresponding string, &quot;abc&quot;.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;aaa&quot;,&quot;bob&quot;,&quot;ccc&quot;,&quot;ddd&quot;]
+<strong>Output:</strong> &quot;bob&quot;
+<strong>Explanation:</strong> All the integer arrays are [0, 0] except for &quot;bob&quot;, which corresponds to [13, -13].
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>3 &lt;= words.length &lt;= 100</code></li>
+	<li><code>n == words[i].length</code></li>
+	<li><code>2 &lt;= n &lt;= 20</code></li>
+	<li><code>words[i]</code> consists of lowercase English letters.</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2400-2499/2452.Words Within Two Edits of Dictionary/README.md

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+# [2452. 距离字典两次编辑以内的单词](https://leetcode.cn/problems/words-within-two-edits-of-dictionary)
+
+[English Version](/solution/2400-2499/2452.Words%20Within%20Two%20Edits%20of%20Dictionary/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你两个字符串数组&nbsp;<code>queries</code> 和&nbsp;<code>dictionary</code>&nbsp;。数组中所有单词都只包含小写英文字母，且长度都相同。</p>
+
+<p>一次 <strong>编辑</strong>&nbsp;中，你可以从 <code>queries</code>&nbsp;中选择一个单词，将任意一个字母修改成任何其他字母。从&nbsp;<code>queries</code>&nbsp;中找到所有满足以下条件的字符串：<strong>不超过</strong>&nbsp;两次编辑内，字符串与&nbsp;<code>dictionary</code>&nbsp;中某个字符串相同。</p>
+
+<p>请你返回<em>&nbsp;</em><code>queries</code>&nbsp;中的单词列表，这些单词距离&nbsp;<code>dictionary</code>&nbsp;中的单词&nbsp;<strong>编辑次数</strong>&nbsp;不超过&nbsp;<strong>两次</strong>&nbsp;。单词返回的顺序需要与&nbsp;<code>queries</code>&nbsp;中原本顺序相同。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><b>输入：</b>queries = ["word","note","ants","wood"], dictionary = ["wood","joke","moat"]
+<b>输出：</b>["word","note","wood"]
+<strong>解释：</strong>
+- 将 "word" 中的 'r' 换成 'o' ，得到 dictionary 中的单词 "wood" 。
+- 将 "note" 中的 'n' 换成 'j' 且将 't' 换成 'k' ，得到 "joke" 。
+- "ants" 需要超过 2 次编辑才能得到 dictionary 中的单词。
+- "wood" 不需要修改（0 次编辑），就得到 dictionary 中相同的单词。
+所以我们返回 ["word","note","wood"] 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><b>输入：</b>queries = ["yes"], dictionary = ["not"]
+<b>输出：</b>[]
+<strong>解释：</strong>
+"yes" 需要超过 2 次编辑才能得到 "not" 。
+所以我们返回空数组。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= queries.length, dictionary.length &lt;= 100</code></li>
+	<li><code>n == queries[i].length == dictionary[j].length</code></li>
+	<li><code>1 &lt;= n &lt;= 100</code></li>
+	<li>所有&nbsp;<code>queries[i]</code> 和&nbsp;<code>dictionary[j]</code>&nbsp;都只包含小写英文字母。</li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2400-2499/2452.Words Within Two Edits of Dictionary/README_EN.md

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+# [2452. Words Within Two Edits of Dictionary](https://leetcode.com/problems/words-within-two-edits-of-dictionary)
+
+[中文文档](/solution/2400-2499/2452.Words%20Within%20Two%20Edits%20of%20Dictionary/README.md)
+
+## Description
+
+<p>You are given two string arrays, <code>queries</code> and <code>dictionary</code>. All words in each array comprise of lowercase English letters and have the same length.</p>
+
+<p>In one <strong>edit</strong> you can take a word from <code>queries</code>, and change any letter in it to any other letter. Find all words from <code>queries</code> that, after a <strong>maximum</strong> of two edits, equal some word from <code>dictionary</code>.</p>
+
+<p>Return<em> a list of all words from </em><code>queries</code><em>, </em><em>that match with some word from </em><code>dictionary</code><em> after a maximum of <strong>two edits</strong></em>. Return the words in the <strong>same order</strong> they appear in <code>queries</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> queries = [&quot;word&quot;,&quot;note&quot;,&quot;ants&quot;,&quot;wood&quot;], dictionary = [&quot;wood&quot;,&quot;joke&quot;,&quot;moat&quot;]
+<strong>Output:</strong> [&quot;word&quot;,&quot;note&quot;,&quot;wood&quot;]
+<strong>Explanation:</strong>
+- Changing the &#39;r&#39; in &quot;word&quot; to &#39;o&#39; allows it to equal the dictionary word &quot;wood&quot;.
+- Changing the &#39;n&#39; to &#39;j&#39; and the &#39;t&#39; to &#39;k&#39; in &quot;note&quot; changes it to &quot;joke&quot;.
+- It would take more than 2 edits for &quot;ants&quot; to equal a dictionary word.
+- &quot;wood&quot; can remain unchanged (0 edits) and match the corresponding dictionary word.
+Thus, we return [&quot;word&quot;,&quot;note&quot;,&quot;wood&quot;].
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> queries = [&quot;yes&quot;], dictionary = [&quot;not&quot;]
+<strong>Output:</strong> []
+<strong>Explanation:</strong>
+Applying any two edits to &quot;yes&quot; cannot make it equal to &quot;not&quot;. Thus, we return an empty array.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= queries.length, dictionary.length &lt;= 100</code></li>
+	<li><code>n == queries[i].length == dictionary[j].length</code></li>
+	<li><code>1 &lt;= n &lt;= 100</code></li>
+	<li>All <code>queries[i]</code> and <code>dictionary[j]</code> are composed of lowercase English letters.</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->