feat: add solutions to lc problems: No.2460~2463

* No.2460.Apply Operations to an Array
* No.2461.Maximum Sum of Distinct Subarrays With Length K
* No.2462.Total Cost to Hire K Workers
* No.2463.Minimum Total Distance Traveled

Author: yanglbme

solution/0000-0099/0005.Longest Palindromic Substring/README_EN.md

@@ -105,7 +105,7 @@ class Solution {
 class Solution {
     private String s;
     private int n;
-    
+
     public String longestPalindrome(String s) {
         this.s = s;
         n = s.length();
@@ -121,7 +121,7 @@ class Solution {
         }
         return s.substring(start, start + mx);
     }
-    
+
     private int f(int l, int r) {
         while (l >= 0 && r < n && s.charAt(l) == s.charAt(r)) {
             --l;

solution/0100-0199/0158.Read N Characters Given read4 II - Call Multiple Times/README_EN.md

@@ -132,7 +132,7 @@ class Solution:
 ```java
 /**
  * The read4 API is defined in the parent class Reader4.
- *     int read4(char[] buf4); 
+ *     int read4(char[] buf4);
  */
 
 public class Solution extends Reader4 {

solution/0100-0199/0158.Read N Characters Given read4 II - Call Multiple Times/Solution.py

@@ -1,6 +1,7 @@
 # The read4 API is already defined for you.
 # def read4(buf4: List[str]) -> int:
 
+
 class Solution:
     def __init__(self):
         self.buf4 = [None] * 4

solution/0200-0299/0254.Factor Combinations/Solution.py

@@ -10,6 +10,7 @@ def dfs(n, i):
                     dfs(n // j, j)
                     t.pop()
                 j += 1
+
         t = []
         ans = []
         dfs(n, 2)

solution/0400-0499/0433.Minimum Genetic Mutation/README_EN.md

@@ -6,49 +6,40 @@
 
 <p>A gene string can be represented by an 8-character long string, with choices from <code>&#39;A&#39;</code>, <code>&#39;C&#39;</code>, <code>&#39;G&#39;</code>, and <code>&#39;T&#39;</code>.</p>
 
-<p>Suppose we need to investigate a mutation from a gene string <code>start</code> to a gene string <code>end</code> where one mutation is defined as one single character changed in the gene string.</p>
+<p>Suppose we need to investigate a mutation from a gene string <code>startGene</code> to a gene string <code>endGene</code> where one mutation is defined as one single character changed in the gene string.</p>
 
 <ul>
 	<li>For example, <code>&quot;AACCGGTT&quot; --&gt; &quot;AACCGGTA&quot;</code> is one mutation.</li>
 </ul>
 
 <p>There is also a gene bank <code>bank</code> that records all the valid gene mutations. A gene must be in <code>bank</code> to make it a valid gene string.</p>
 
-<p>Given the two gene strings <code>start</code> and <code>end</code> and the gene bank <code>bank</code>, return <em>the minimum number of mutations needed to mutate from </em><code>start</code><em> to </em><code>end</code>. If there is no such a mutation, return <code>-1</code>.</p>
+<p>Given the two gene strings <code>startGene</code> and <code>endGene</code> and the gene bank <code>bank</code>, return <em>the minimum number of mutations needed to mutate from </em><code>startGene</code><em> to </em><code>endGene</code>. If there is no such a mutation, return <code>-1</code>.</p>
 
 <p>Note that the starting point is assumed to be valid, so it might not be included in the bank.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 
 <pre>
-<strong>Input:</strong> start = &quot;AACCGGTT&quot;, end = &quot;AACCGGTA&quot;, bank = [&quot;AACCGGTA&quot;]
+<strong>Input:</strong> startGene = &quot;AACCGGTT&quot;, endGene = &quot;AACCGGTA&quot;, bank = [&quot;AACCGGTA&quot;]
 <strong>Output:</strong> 1
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
 
 <pre>
-<strong>Input:</strong> start = &quot;AACCGGTT&quot;, end = &quot;AAACGGTA&quot;, bank = [&quot;AACCGGTA&quot;,&quot;AACCGCTA&quot;,&quot;AAACGGTA&quot;]
+<strong>Input:</strong> startGene = &quot;AACCGGTT&quot;, endGene = &quot;AAACGGTA&quot;, bank = [&quot;AACCGGTA&quot;,&quot;AACCGCTA&quot;,&quot;AAACGGTA&quot;]
 <strong>Output:</strong> 2
 </pre>
 
-<p><strong class="example">Example 3:</strong></p>
-
-<pre>
-<strong>Input:</strong> start = &quot;AAAAACCC&quot;, end = &quot;AACCCCCC&quot;, bank = [&quot;AAAACCCC&quot;,&quot;AAACCCCC&quot;,&quot;AACCCCCC&quot;]
-<strong>Output:</strong> 3
-</pre>
-
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>start.length == 8</code></li>
-	<li><code>end.length == 8</code></li>
 	<li><code>0 &lt;= bank.length &lt;= 10</code></li>
-	<li><code>bank[i].length == 8</code></li>
-	<li><code>start</code>, <code>end</code>, and <code>bank[i]</code> consist of only the characters <code>[&#39;A&#39;, &#39;C&#39;, &#39;G&#39;, &#39;T&#39;]</code>.</li>
+	<li><code>startGene.length == endGene.length == bank[i].length == 8</code></li>
+	<li><code>startGene</code>, <code>endGene</code>, and <code>bank[i]</code> consist of only the characters <code>[&#39;A&#39;, &#39;C&#39;, &#39;G&#39;, &#39;T&#39;]</code>.</li>
 </ul>
 
 ## Solutions

solution/0700-0799/0769.Max Chunks To Make Sorted/README.md

@@ -32,6 +32,7 @@
 <strong>解释:</strong>
 我们可以把它分成两块，例如 [1, 0], [2, 3, 4]。
 然而，分成 [1, 0], [2], [3], [4] 可以得到最多的块数。
+对每个块单独排序后，结果为 [0, 1], [2], [3], [4]
 </pre>
 
 <p>&nbsp;</p>

solution/0700-0799/0779.K-th Symbol in Grammar/README.md

@@ -30,7 +30,7 @@
 <strong>输出:</strong> 0
 <strong>解释:</strong> 
 第一行: 0 
-第二行: 0<u>1</u>
+第二行: <u>0</u>1
 </pre>
 
 <p><strong>示例 3:</strong></p>

solution/0900-0999/0921.Minimum Add to Make Parentheses Valid/README.md

@@ -14,7 +14,7 @@
 	<li>它可以被写作&nbsp;<code>(A)</code>，其中&nbsp;<code>A</code>&nbsp;是有效字符串。</li>
 </ul>
 
-<p>给定一个括号字符串 <code>s</code> ，移动N次，你就可以在字符串的任何位置插入一个括号。</p>
+<p>给定一个括号字符串 <code>s</code> ，在每一次操作中，你都可以在字符串的任何位置插入一个括号</p>
 
 <ul>
 	<li>例如，如果 <code>s = "()))"</code> ，你可以插入一个开始括号为 <code>"(()))"</code> 或结束括号为 <code>"())))"</code> 。</li>

solution/1300-1399/1367.Linked List in Binary Tree/README.md

@@ -1,4 +1,4 @@
-# [1367. 二叉树中的列表](https://leetcode.cn/problems/linked-list-in-binary-tree)
+# [1367. 二叉树中的链表](https://leetcode.cn/problems/linked-list-in-binary-tree)
 
 [English Version](/solution/1300-1399/1367.Linked%20List%20in%20Binary%20Tree/README_EN.md)
 

solution/1700-1799/1784.Check if Binary String Has at Most One Segment of Ones/README.md

@@ -10,8 +10,6 @@
 
 <p>如果 <code>s</code> 包含 <strong>零个或一个由连续的 <code>'1'</code> 组成的字段</strong> ，返回 <code>true</code>​​​ 。否则，返回 <code>false</code> 。</p>
 
-<p>如果 <code>s</code>&nbsp;中&nbsp;<strong>由连续若干个&nbsp;<code>'1'</code> 组成的字段</strong>&nbsp;数量不超过 <code>1</code>，返回 <code>true</code>​​​ 。否则，返回 <code>false</code> 。</p>
-
 <p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>