feat: add new lc problems and solutions (doocs#2541)

Author: yanglbme

package-lock.json

@@ -10,31 +10,29 @@
 
 <p>顾客表：<code>Customers</code></p>
 
-<pre>
-+---------------+---------+
+<pre>+---------------+---------+
 | Column Name   | Type    |
 +---------------+---------+
 | customer_id   | int     |
 | customer_name | varchar |
 | email         | varchar |
 +---------------+---------+
-customer_id 是这张表具有唯一值的列。
+customer_id 是这张表的主键。
 此表的每一行包含了某在线商店顾客的姓名和电子邮件。
 </pre>
 
 <p>&nbsp;</p>
 
 <p>联系方式表：<code>Contacts</code></p>
 
-<pre>
-+---------------+---------+
+<pre>+---------------+---------+
 | Column Name   | Type    |
 +---------------+---------+
 | user_id       | id      |
 | contact_name  | varchar |
 | contact_email | varchar |
 +---------------+---------+
-(user_id, contact_email) 是这张表的主键（具有唯一值的列的组合）。
+(user_id, contact_email) 是这张表的主键。
 此表的每一行表示编号为 user_id 的顾客的某位联系人的姓名和电子邮件。
 此表包含每位顾客的联系人信息，但顾客的联系人不一定存在于顾客表中。
 </pre>
@@ -43,40 +41,33 @@ customer_id 是这张表具有唯一值的列。
 
 <p>发票表：<code>Invoices</code></p>
 
-<pre>
-+--------------+---------+
+<pre>+--------------+---------+
 | Column Name  | Type    |
 +--------------+---------+
 | invoice_id   | int     |
 | price        | int     |
 | user_id      | int     |
 +--------------+---------+
-invoice_id 是这张表具有唯一值的列。
+invoice_id 是这张表的主键。
 此表的每一行分别表示编号为 user_id 的顾客拥有有一张编号为 invoice_id、价格为 price 的发票。
 </pre>
 
 <p>&nbsp;</p>
 
-<p>为每张发票 <code>invoice_id</code> 编写一个查询方案以查找以下内容：</p>
+<p>为每张发票 <code>invoice_id</code> 编写一个SQL查询以查找以下内容：</p>
 
 <ul>
 	<li><code>customer_name</code>：与发票相关的顾客名称。</li>
 	<li><code>price</code>：发票的价格。</li>
-	<li><code>contacts_cnt</code>：该顾客的联系人数量</li>
-	<li><code>trusted_contacts_cnt</code>：可信联系人的数量：既是该顾客的联系人又是商店顾客的联系人数量（即：可信联系人的电子邮件存在于 <meta charset="UTF-8" />&nbsp;<code>Customers</code>&nbsp;表中）。</li>
+	<li><code>contacts_cnt</code>：该顾客的联系人数量。</li>
+	<li><code>trusted_contacts_cnt</code>：可信联系人的数量：既是该顾客的联系人又是商店顾客的联系人数量（即：可信联系人的电子邮件存在于客户表中）。</li>
 </ul>
 
-<p>返回结果按照&nbsp;<code>invoice_id</code>&nbsp;<strong>排序</strong>。</p>
-
-<p>结果的格式如下例所示。</p>
-
-<p>&nbsp;</p>
+<p>将查询的结果按照&nbsp;<code>invoice_id</code>&nbsp;排序。</p>
 
-<p><strong>示例 1：</strong></p>
+<p>查询结果的格式如下例所示：</p>
 
-<pre>
-<strong>输入：</strong>
-<code>Customers</code> table:
+<pre><code>Customers</code> table:
 +-------------+---------------+--------------------+
 | customer_id | customer_name | email              |
 +-------------+---------------+--------------------+
@@ -107,7 +98,7 @@ Invoices table:
 | 55         | 500   | 13      |
 | 44         | 60    | 6       |
 +------------+-------+---------+
-<strong>输出：</strong>
+Result table:
 +------------+---------------+-------+--------------+----------------------+
 | invoice_id | customer_name | price | contacts_cnt | trusted_contacts_cnt |
 +------------+---------------+-------+--------------+----------------------+
@@ -118,7 +109,6 @@ Invoices table:
 | 88         | Alice         | 200   | 3            | 2                    |
 | 99         | Bob           | 300   | 2            | 0                    |
 +------------+---------------+-------+--------------+----------------------+
-<strong>解释：</strong>
 Alice 有三位联系人，其中两位(Bob 和 John)是可信联系人。
 Bob 有两位联系人, 他们中的任何一位都不是可信联系人。
 Alex 只有一位联系人(Alice)，并是一位可信联系人。

solution/1300-1399/1364.Number of Trusted Contacts of a Customer/README.md

@@ -16,7 +16,7 @@
 | customer_name | varchar |
 | email         | varchar |
 +---------------+---------+
-customer_id is the column of unique values for this table.
+customer_id is the primary key for this table.
 Each row of this table contains the name and the email of a customer of an online shop.
 </pre>
 
@@ -32,7 +32,7 @@ Each row of this table contains the name and the email of a customer of an onlin
 | contact_name  | varchar |
 | contact_email | varchar |
 +---------------+---------+
-(user_id, contact_email) is the primary key (combination of columns with unique values) for this table.
+(user_id, contact_email) is the primary key for this table.
 Each row of this table contains the name and email of one contact of customer with user_id.
 This table contains information about people each customer trust. The contact may or may not exist in the Customers table.
 </pre>
@@ -49,13 +49,13 @@ This table contains information about people each customer trust. The contact ma
 | price        | int     |
 | user_id      | int     |
 +--------------+---------+
-invoice_id is the column of unique values for this table.
+invoice_id is the primary key for this table.
 Each row of this table indicates that user_id has an invoice with invoice_id and a price.
 </pre>
 
 <p>&nbsp;</p>
 
-<p>Write a solution to find the following for each <code>invoice_id</code>:</p>
+<p>Write an SQL query to find the following for each <code>invoice_id</code>:</p>
 
 <ul>
 	<li><code>customer_name</code>: The name of the customer the invoice is related to.</li>
@@ -66,7 +66,7 @@ Each row of this table indicates that user_id has an invoice with invoice_id and
 
 <p>Return the result table <strong>ordered</strong> by <code>invoice_id</code>.</p>
 
-<p>The result format is in the following example.</p>
+<p>The query result format is in the following example.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>

solution/1300-1399/1364.Number of Trusted Contacts of a Customer/README_EN.md

@@ -2,7 +2,7 @@
 
 [English Version](/solution/1600-1699/1668.Maximum%20Repeating%20Substring/README_EN.md)
 
-<!-- tags:字符串,字符串匹配 -->
+<!-- tags:字符串,动态规划,字符串匹配 -->
 
 ## 题目描述
 

solution/1600-1699/1668.Maximum Repeating Substring/README.md

@@ -2,7 +2,7 @@
 
 [中文文档](/solution/1600-1699/1668.Maximum%20Repeating%20Substring/README.md)
 
-<!-- tags:String,String Matching -->
+<!-- tags:String,Dynamic Programming,String Matching -->
 
 ## Description
 

solution/1600-1699/1668.Maximum Repeating Substring/README_EN.md

@@ -2,7 +2,7 @@
 
 [English Version](/solution/1700-1799/1738.Find%20Kth%20Largest%20XOR%20Coordinate%20Value/README_EN.md)
 
-<!-- tags:位运算,数组,分治,矩阵,前缀和,快速选择,堆（优先队列） -->
+<!-- tags:位运算,数组,分治,矩阵,前缀和,快速选择,排序,堆（优先队列） -->
 
 ## 题目描述
 

solution/1700-1799/1738.Find Kth Largest XOR Coordinate Value/README.md

@@ -2,7 +2,7 @@
 
 [中文文档](/solution/1700-1799/1738.Find%20Kth%20Largest%20XOR%20Coordinate%20Value/README.md)
 
-<!-- tags:Bit Manipulation,Array,Divide and Conquer,Matrix,Prefix Sum,Quickselect,Heap (Priority Queue) -->
+<!-- tags:Bit Manipulation,Array,Divide and Conquer,Matrix,Prefix Sum,Quickselect,Sorting,Heap (Priority Queue) -->
 
 ## Description
 

solution/1700-1799/1738.Find Kth Largest XOR Coordinate Value/README_EN.md

@@ -2,7 +2,7 @@
 
 [English Version](/solution/2000-2099/2009.Minimum%20Number%20of%20Operations%20to%20Make%20Array%20Continuous/README_EN.md)
 
-<!-- tags:数组,二分查找 -->
+<!-- tags:数组,哈希表,二分查找,滑动窗口 -->
 
 ## 题目描述
 

solution/2000-2099/2009.Minimum Number of Operations to Make Array Continuous/README.md

@@ -2,7 +2,7 @@
 
 [中文文档](/solution/2000-2099/2009.Minimum%20Number%20of%20Operations%20to%20Make%20Array%20Continuous/README.md)
 
-<!-- tags:Array,Binary Search -->
+<!-- tags:Array,Hash Table,Binary Search,Sliding Window -->
 
 ## Description
 

solution/2000-2099/2009.Minimum Number of Operations to Make Array Continuous/README_EN.md

@@ -2,7 +2,7 @@
 
 [English Version](/solution/2600-2699/2681.Power%20of%20Heroes/README_EN.md)
 
-<!-- tags:数组,数学,前缀和,排序 -->
+<!-- tags:数组,数学,动态规划,前缀和,排序 -->
 
 ## 题目描述
 

solution/2600-2699/2681.Power of Heroes/README.md

@@ -2,7 +2,7 @@
 
 [中文文档](/solution/2600-2699/2681.Power%20of%20Heroes/README.md)
 
-<!-- tags:Array,Math,Prefix Sum,Sorting -->
+<!-- tags:Array,Math,Dynamic Programming,Prefix Sum,Sorting -->
 
 ## Description
 

solution/2600-2699/2681.Power of Heroes/README_EN.md

@@ -6,61 +6,44 @@
 
 ## Description
 
-<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code>.<br />
-<br />
-The numbers from <code>0</code> to <code>n - 1</code> are divided into three groups numbered from <code>1</code> to <code>3</code>, where number <code>i</code> belongs to group <code>nums[i]</code>. Notice that some groups may be <strong>empty</strong>.<br />
-<br />
-You are allowed to perform this operation any number of times:</p>
+<p>You are given an integer array <code>nums</code>. Each element in <code>nums</code> is 1, 2 or 3. In each operation, you can remove an element from&nbsp;<code>nums</code>. Return the <strong>minimum</strong> number of operations to make <code>nums</code> <strong>non-decreasing</strong>.</p>
 
-<ul>
-	<li>Pick number <code>x</code> and change its group. More formally, change <code>nums[x]</code> to any number from <code>1</code> to <code>3</code>.</li>
-</ul>
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
 
-<p>A new array <code>res</code> is constructed using the following procedure:</p>
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [2,1,3,2,1]</span></p>
 
-<ol>
-	<li>Sort the numbers in each group independently.</li>
-	<li>Append the elements of groups <code>1</code>, <code>2</code>, and <code>3</code> to <code>res</code> <strong>in this order</strong>.</li>
-</ol>
+<p><strong>Output:</strong> <span class="example-io">3</span></p>
 
-<p>Array <code>nums</code> is called a <strong>beautiful array</strong> if the constructed array <code>res</code> is sorted in <strong>non-decreasing</strong> order.</p>
+<p><strong>Explanation:</strong></p>
 
-<p>Return <em>the <strong>minimum</strong> number of operations to make </em><code>nums</code><em> a <strong>beautiful array</strong></em>.</p>
+<p>One of the optimal solutions is to remove <code>nums[0]</code>, <code>nums[2]</code> and <code>nums[3]</code>.</p>
+</div>
 
-<p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+<p><strong class="example">Example 2:</strong></p>
 
-<pre>
-<strong>Input:</strong> nums = [2,1,3,2,1]
-<strong>Output:</strong> 3
-<strong>Explanation:</strong> It&#39;s optimal to perform three operations:
-1. change nums[0] to 1.
-2. change nums[2] to 1.
-3. change nums[3] to 1.
-After performing the operations and sorting the numbers in each group, group 1 becomes equal to [0,1,2,3,4] and group 2 and group 3 become empty. Hence, res is equal to [0,1,2,3,4] which is sorted in non-decreasing order.
-It can be proven that there is no valid sequence of less than three operations.
-</pre>
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1,3,2,1,3,3]</span></p>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><strong>Output:</strong> <span class="example-io">2</span></p>
+
+<p><strong>Explanation:</strong></p>
 
-<pre>
-<strong>Input:</strong> nums = [1,3,2,1,3,3]
-<strong>Output:</strong> 2
-<strong>Explanation:</strong> It&#39;s optimal to perform two operations:
-1. change nums[1] to 1.
-2. change nums[2] to 1.
-After performing the operations and sorting the numbers in each group, group 1 becomes equal to [0,1,2,3], group 2 becomes empty, and group 3 becomes equal to [4,5]. Hence, res is equal to [0,1,2,3,4,5] which is sorted in non-decreasing order.
-It can be proven that there is no valid sequence of less than two operations.
-</pre>
+<p>One of the optimal solutions is to remove <code>nums[1]</code> and <code>nums[2]</code>.</p>
+</div>
 
 <p><strong class="example">Example 3:</strong></p>
 
-<pre>
-<strong>Input:</strong> nums = [2,2,2,2,3,3]
-<strong>Output:</strong> 0
-<strong>Explanation:</strong> It&#39;s optimal to not perform operations.
-After sorting the numbers in each group, group 1 becomes empty, group 2 becomes equal to [0,1,2,3] and group 3 becomes equal to [4,5]. Hence, res is equal to [0,1,2,3,4,5] which is sorted in non-decreasing order.
-</pre>
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [2,2,2,2,3,3]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">0</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p><code>nums</code> is already non-decreasing.</p>
+</div>
 
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
@@ -70,6 +53,9 @@ After sorting the numbers in each group, group 1 becomes empty, group 2 becomes
 	<li><code>1 &lt;= nums[i] &lt;= 3</code></li>
 </ul>
 
+<p>&nbsp;</p>
+<strong>Follow-up:</strong> Can you come up with an algorithm that runs in <code>O(n)</code> time complexity?
+
 ## Solutions
 
 ### Solution 1: Dynamic Programming