feat: add solutions to lc problem: No.3172

solution/0100-0199/0152.Maximum Product Subarray/README.md

@@ -23,15 +23,15 @@ tags:
 
 <p>&nbsp;</p>
 
-<p><strong>示例 1:</strong></p>
+<p><strong class="example">示例 1:</strong></p>
 
 <pre>
 <strong>输入:</strong> nums = [2,3,-2,4]
 <strong>输出:</strong> <code>6</code>
 <strong>解释:</strong>&nbsp;子数组 [2,3] 有最大乘积 6。
 </pre>
 
-<p><strong>示例 2:</strong></p>
+<p><strong class="example">示例 2:</strong></p>
 
 <pre>
 <strong>输入:</strong> nums = [-2,0,-1]

solution/3100-3199/3172.Second Day Verification/README.md

@@ -0,0 +1,133 @@
+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3172.Second%20Day%20Verification/README.md
+---
+
+<!-- problem:start -->
+
+# [3172. 第二天验证 🔒](https://leetcode.cn/problems/second-day-verification)
+
+[English Version](/solution/3100-3199/3172.Second%20Day%20Verification/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>表：<code>emails</code></p>
+
+<pre>
++-------------+----------+
+| Column Name | Type     | 
++-------------+----------+
+| email_id    | int      |
+| user_id     | int      |
+| signup_date | datetime |
++-------------+----------+
+(email_id, user_id) 是这张表的主键（有不同值的列的组合）。
+这张表的每一行包含 email ID，user ID 和注册日期。
+</pre>
+
+<p>表：<code>texts</code></p>
+
+<pre>
++---------------+----------+
+| Column Name   | Type     | 
++---------------+----------+
+| text_id       | int      |
+| email_id      | int      |
+| signup_action | enum     |
+| action_date   | datetime |
++---------------+----------+
+(text_id, email_id) 是这张表的主键（有不同值的列的组合）。
+signup_action 是 ('Verified', 'Not Verified') 的枚举类型。
+这张表的每一行包含 text ID，email ID，注册操作和操作日期。
+</pre>
+
+<p>编写一个解决方案来找到&nbsp;<strong>第二天验证注册</strong>&nbsp;的用户 ID。</p>
+
+<p>返回结果表以&nbsp;<code>user_id</code> <strong>升序&nbsp;</strong>排序。</p>
+
+<p>结果格式如下所示。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例：</strong></p>
+
+<div class="example-block">
+<p><b>输入：</b></p>
+
+<p>emails 表：</p>
+
+<pre class="example-io">
++----------+---------+---------------------+
+| email_id | user_id | signup_date         |
++----------+---------+---------------------+
+| 125      | 7771    | 2022-06-14 09:30:00|
+| 433      | 1052    | 2022-07-09 08:15:00|
+| 234      | 7005    | 2022-08-20 10:00:00|
++----------+---------+---------------------+
+</pre>
+
+<p>texts 表：</p>
+
+<pre class="example-io">
++---------+----------+--------------+---------------------+
+| text_id | email_id | signup_action| action_date         |
++---------+----------+--------------+---------------------+
+| 1       | 125      | Verified     | 2022-06-15 08:30:00|
+| 2       | 433      | Not Verified | 2022-07-10 10:45:00|
+| 4       | 234      | Verified     | 2022-08-21 09:30:00|
++---------+----------+--------------+---------------------+
+    </pre>
+
+<p><strong>输出：</strong></p>
+
+<pre class="example-io">
++---------+
+| user_id |
++---------+
+| 7005    |
+| 7771    |
++---------+
+</pre>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li>email_id 为 7005 的用户在 2022-08-20 10:00:00 注册并且在第二天验证。</li>
+	<li>email_id 为 7771 的用户在 2022-06-14 09:30:00 注册并且在第二天验证。</li>
+</ul>
+</div>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：双表关联
+
+我们可以通过内连接两个表，然后根据 `DATEDIFF` 函数计算出注册日期和操作日期的差值是否等于 1，以及注册操作是否为 `Verified`，来筛选出满足条件的用户 ID。
+
+<!-- tabs:start -->
+
+#### MySQL
+
+```sql
+# Write your MySQL query statement below
+SELECT user_id
+FROM
+    Emails AS e
+    JOIN texts AS t
+        ON e.email_id = t.email_id
+        AND DATEDIFF(action_date, signup_date) = 1
+        AND signup_action = 'Verified'
+ORDER BY 1;
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3100-3199/3172.Second Day Verification/README_EN.md

@@ -0,0 +1,132 @@
+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3172.Second%20Day%20Verification/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3172. Second Day Verification 🔒](https://leetcode.com/problems/second-day-verification)
+
+[中文文档](/solution/3100-3199/3172.Second%20Day%20Verification/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>Table: <code>emails</code></p>
+
+<pre>
++-------------+----------+
+| Column Name | Type     | 
++-------------+----------+
+| email_id    | int      |
+| user_id     | int      |
+| signup_date | datetime |
++-------------+----------+
+(email_id, user_id) is the primary key (combination of columns with unique values) for this table.
+Each row of this table contains the email ID, user ID, and signup date.
+</pre>
+
+<p>Table: <code>texts</code></p>
+
+<pre>
++---------------+----------+
+| Column Name   | Type     | 
++---------------+----------+
+| text_id       | int      |
+| email_id      | int      |
+| signup_action | enum     |
+| action_date   | datetime |
++---------------+----------+
+(text_id, email_id) is the primary key (combination of columns with unique values) for this table. 
+signup_action is an enum type of (&#39;Verified&#39;, &#39;Not Verified&#39;). 
+Each row of this table contains the text ID, email ID, signup action, and action date.
+</pre>
+
+<p>Write a Solution to find the user IDs of those who <strong>verified</strong> their <strong>sign-up</strong> on the <strong>second day</strong>.</p>
+
+<p>Return <em>the result table ordered by</em> <code>user_id</code> <em>in <strong>ascending</strong> order</em>.</p>
+
+<p>The result format is in the following example.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong></p>
+
+<p>emails table:</p>
+
+<pre class="example-io">
++----------+---------+---------------------+
+| email_id | user_id | signup_date         |
++----------+---------+---------------------+
+| 125      | 7771    | 2022-06-14 09:30:00|
+| 433      | 1052    | 2022-07-09 08:15:00|
+| 234      | 7005    | 2022-08-20 10:00:00|
++----------+---------+---------------------+
+</pre>
+
+<p>texts table:</p>
+
+<pre class="example-io">
++---------+----------+--------------+---------------------+
+| text_id | email_id | signup_action| action_date         |
++---------+----------+--------------+---------------------+
+| 1       | 125      | Verified     | 2022-06-15 08:30:00|
+| 2       | 433      | Not Verified | 2022-07-10 10:45:00|
+| 4       | 234      | Verified     | 2022-08-21 09:30:00|
++---------+----------+--------------+---------------------+
+    </pre>
+
+<p><strong>Output:</strong></p>
+
+<pre class="example-io">
++---------+
+| user_id |
++---------+
+| 7005    |
+| 7771    |
++---------+
+</pre>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>User with email_id 7005 signed up on 2022-08-20 10:00:00 and&nbsp;verified on second day of the signup.</li>
+	<li>User with email_id 7771 signed up on 2022-06-14 09:30:00 and&nbsp;verified on second day of the signup.</li>
+</ul>
+</div>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Joining Two Tables
+
+We can join the two tables and then use the `DATEDIFF` function to calculate whether the difference between the registration date and the operation date is equal to 1, and whether the registration operation is `Verified`, to filter out the user IDs that meet the conditions.
+
+<!-- tabs:start -->
+
+#### MySQL
+
+```sql
+# Write your MySQL query statement below
+SELECT user_id
+FROM
+    Emails AS e
+    JOIN texts AS t
+        ON e.email_id = t.email_id
+        AND DATEDIFF(action_date, signup_date) = 1
+        AND signup_action = 'Verified'
+ORDER BY 1;
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3100-3199/3172.Second Day Verification/Solution.sql

@@ -0,0 +1,9 @@
+# Write your MySQL query statement below
+SELECT user_id
+FROM
+    Emails AS e
+    JOIN texts AS t
+        ON e.email_id = t.email_id
+        AND DATEDIFF(action_date, signup_date) = 1
+        AND signup_action = 'Verified'
+ORDER BY 1;

solution/DATABASE_README.md

@@ -281,6 +281,7 @@
 | 3150 | [无效的推文 II](/solution/3100-3199/3150.Invalid%20Tweets%20II/README.md)                                                                                    | `数据库` | 简单 | 🔒   |
 | 3156 | [员工任务持续时间和并发任务](/solution/3100-3199/3156.Employee%20Task%20Duration%20and%20Concurrent%20Tasks/README.md)                                       | `数据库` | 困难 | 🔒   |
 | 3166 | [计算停车费与时长](/solution/3100-3199/3166.Calculate%20Parking%20Fees%20and%20Duration/README.md)                                                           | `数据库` | 中等 | 🔒   |
+| 3172 | [第二天验证](/solution/3100-3199/3172.Second%20Day%20Verification/README.md)                                                                                 |          | 简单 | 🔒   |
 
 ## 版权
 

solution/DATABASE_README_EN.md

@@ -279,6 +279,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
 | 3150 | [Invalid Tweets II](/solution/3100-3199/3150.Invalid%20Tweets%20II/README_EN.md)                                                                                                             | `Database` | Easy       | 🔒     |
 | 3156 | [Employee Task Duration and Concurrent Tasks](/solution/3100-3199/3156.Employee%20Task%20Duration%20and%20Concurrent%20Tasks/README_EN.md)                                                   | `Database` | Hard       | 🔒     |
 | 3166 | [Calculate Parking Fees and Duration](/solution/3100-3199/3166.Calculate%20Parking%20Fees%20and%20Duration/README_EN.md)                                                                     | `Database` | Medium     | 🔒     |
+| 3172 | [Second Day Verification](/solution/3100-3199/3172.Second%20Day%20Verification/README_EN.md)                                                                                                 |            | Easy       | 🔒     |
 
 ## Copyright
 

solution/README.md

@@ -3182,6 +3182,7 @@
 |  3169  |  [无需开会的工作日](/solution/3100-3199/3169.Count%20Days%20Without%20Meetings/README.md)  |    |  中等  |  第 400 场周赛  |
 |  3170  |  [删除星号以后字典序最小的字符串](/solution/3100-3199/3170.Lexicographically%20Minimum%20String%20After%20Removing%20Stars/README.md)  |    |  中等  |  第 400 场周赛  |
 |  3171  |  [找到按位与最接近 K 的子数组](/solution/3100-3199/3171.Find%20Subarray%20With%20Bitwise%20AND%20Closest%20to%20K/README.md)  |    |  困难  |  第 400 场周赛  |
+|  3172  |  [第二天验证](/solution/3100-3199/3172.Second%20Day%20Verification/README.md)  |    |  简单  |  🔒  |
 
 ## 版权
 

solution/README_EN.md

@@ -3180,6 +3180,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
 |  3169  |  [Count Days Without Meetings](/solution/3100-3199/3169.Count%20Days%20Without%20Meetings/README_EN.md)  |    |  Medium  |  Weekly Contest 400  |
 |  3170  |  [Lexicographically Minimum String After Removing Stars](/solution/3100-3199/3170.Lexicographically%20Minimum%20String%20After%20Removing%20Stars/README_EN.md)  |    |  Medium  |  Weekly Contest 400  |
 |  3171  |  [Find Subarray With Bitwise AND Closest to K](/solution/3100-3199/3171.Find%20Subarray%20With%20Bitwise%20AND%20Closest%20to%20K/README_EN.md)  |    |  Hard  |  Weekly Contest 400  |
+|  3172  |  [Second Day Verification](/solution/3100-3199/3172.Second%20Day%20Verification/README_EN.md)  |    |  Easy  |  🔒  |
 
 ## Copyright