feat: update lc problems

Author: yanglbme

solution/0200-0299/0262.Trips and Users/README.md

@@ -61,7 +61,7 @@ banned 是一个表示用户是否被禁止的枚举类型，枚举成员为 (
 
 <p><strong>取消率</strong> 的计算方式如下：(被司机或乘客取消的非禁止用户生成的订单数量) / (非禁止用户生成的订单总数)。</p>
 
-<p>编写解决方案找出&nbsp;<code>"2013-10-01"</code><strong>&nbsp;</strong>至&nbsp;<code>"2013-10-03"</code><strong>&nbsp;</strong>期间非禁止用户（<strong>乘客和司机都必须未被禁止</strong>）的取消率。非禁止用户即 banned 为 No 的用户，禁止用户即 banned 为 Yes 的用户。其中取消率 <code>Cancellation Rate</code> 需要四舍五入保留 <strong>两位小数</strong> 。</p>
+<p>编写解决方案找出&nbsp;<code>"2013-10-01"</code><strong>&nbsp;</strong>至&nbsp;<code>"2013-10-03"</code><strong>&nbsp;</strong>期间有&nbsp;<strong>至少&nbsp;</strong>一次行程的非禁止用户（<strong>乘客和司机都必须未被禁止</strong>）的 <strong>取消率</strong>。非禁止用户即 banned 为 No 的用户，禁止用户即 banned 为 Yes 的用户。其中取消率 <code>Cancellation Rate</code> 需要四舍五入保留 <strong>两位小数</strong> 。</p>
 
 <p>返回结果表中的数据 <strong>无顺序要求</strong> 。</p>
 

solution/1300-1399/1363.Largest Multiple of Three/README.md

@@ -7,7 +7,9 @@ source: 第 177 场周赛 Q4
 tags:
     - 贪心
     - 数组
+    - 数学
     - 动态规划
+    - 排序
 ---
 
 <!-- problem:start -->

solution/1300-1399/1363.Largest Multiple of Three/README_EN.md

@@ -7,7 +7,9 @@ source: Weekly Contest 177 Q4
 tags:
     - Greedy
     - Array
+    - Math
     - Dynamic Programming
+    - Sorting
 ---
 
 <!-- problem:start -->

solution/1300-1399/1365.How Many Numbers Are Smaller Than the Current Number/README.md

@@ -7,7 +7,7 @@ source: 第 178 场周赛 Q1
 tags:
     - 数组
     - 哈希表
-    - 计数
+    - 计数排序
     - 排序
 ---
 

solution/1300-1399/1365.How Many Numbers Are Smaller Than the Current Number/README_EN.md

@@ -7,7 +7,7 @@ source: Weekly Contest 178 Q1
 tags:
     - Array
     - Hash Table
-    - Counting
+    - Counting Sort
     - Sorting
 ---
 

solution/1400-1499/1434.Number of Ways to Wear Different Hats to Each Other/README.md

@@ -57,13 +57,6 @@ tags:
 (1,2,3,4) 4 个帽子的排列方案数为 24 。
 </pre>
 
-<p><strong>示例 4：</strong></p>
-
-<pre>
-<strong>输入：</strong>hats = [[1,2,3],[2,3,5,6],[1,3,7,9],[1,8,9],[2,5,7]]
-<strong>输出：</strong>111
-</pre>
-
 <p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>

solution/1400-1499/1434.Number of Ways to Wear Different Hats to Each Other/README_EN.md

@@ -25,7 +25,7 @@ tags:
 
 <p>Given a 2D integer array <code>hats</code>, where <code>hats[i]</code> is a list of all hats preferred by the <code>i<sup>th</sup></code> person.</p>
 
-<p>Return <em>the number of ways that the <code>n</code> people wear different hats to each other</em>.</p>
+<p>Return the number of ways that <code>n</code> people can wear <strong>different</strong> hats from each other.</p>
 
 <p>Since the answer may be too large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>
 

solution/1900-1999/1968.Array With Elements Not Equal to Average of Neighbors/README_EN.md

@@ -47,7 +47,7 @@ When i=3, nums[i] = 5, and the average of its neighbors is (4+3) / 2 = 3.5.
 When i=1, nums[i] = 7, and the average of its neighbors is (9+6) / 2 = 7.5.
 When i=2, nums[i] = 6, and the average of its neighbors is (7+2) / 2 = 4.5.
 When i=3, nums[i] = 2, and the average of its neighbors is (6+0) / 2 = 3.
-</pre>
+Note that the original array [6,2,0,9,7] also satisfies the conditions.</pre>
 
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>

solution/1900-1999/1974.Minimum Time to Type Word Using Special Typewriter/README_EN.md

@@ -36,7 +36,7 @@ tags:
 <pre>
 <strong>Input:</strong> word = &quot;abc&quot;
 <strong>Output:</strong> 5
-<strong>Explanation:
+<strong>Explanation: 
 </strong>The characters are printed as follows:
 - Type the character &#39;a&#39; in 1 second since the pointer is initially on &#39;a&#39;.
 - Move the pointer clockwise to &#39;b&#39; in 1 second.

solution/2100-2199/2138.Divide a String Into Groups of Size k/README_EN.md

@@ -22,7 +22,7 @@ tags:
 <p>A string <code>s</code> can be partitioned into groups of size <code>k</code> using the following procedure:</p>
 
 <ul>
-	<li>The first group consists of the first <code>k</code> characters of the string, the second group consists of the next <code>k</code> characters of the string, and so on. Each character can be a part of <strong>exactly one</strong> group.</li>
+	<li>The first group consists of the first <code>k</code> characters of the string, the second group consists of the next <code>k</code> characters of the string, and so on. Each element can be a part of <strong>exactly one</strong> group.</li>
 	<li>For the last group, if the string <strong>does not</strong> have <code>k</code> characters remaining, a character <code>fill</code> is used to complete the group.</li>
 </ul>