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feat: add solutions to lc problem: No.3172 (#3000)
No.3172.Second Day Verification
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Diff for: ‎solution/0100-0199/0152.Maximum Product Subarray/README.md

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

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

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---
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comments: true
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difficulty: 简单
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3172.Second%20Day%20Verification/README.md
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---
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<!-- problem:start -->
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# [3172. 第二天验证 🔒](https://leetcode.cn/problems/second-day-verification)
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[English Version](/solution/3100-3199/3172.Second%20Day%20Verification/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>表：<code>emails</code></p>
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<pre>
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+-------------+----------+
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| Column Name | Type |
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+-------------+----------+
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| email_id | int |
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| user_id | int |
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| signup_date | datetime |
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+-------------+----------+
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(email_id, user_id) 是这张表的主键（有不同值的列的组合）。
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这张表的每一行包含 email ID，user ID 和注册日期。
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</pre>
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<p>表：<code>texts</code></p>
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<pre>
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+---------------+----------+
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| Column Name | Type |
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+---------------+----------+
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| text_id | int |
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| email_id | int |
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| signup_action | enum |
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| action_date | datetime |
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+---------------+----------+
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(text_id, email_id) 是这张表的主键（有不同值的列的组合）。
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signup_action 是 ('Verified', 'Not Verified') 的枚举类型。
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这张表的每一行包含 text ID，email ID，注册操作和操作日期。
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</pre>
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<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>
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<p>返回结果表以&nbsp;<code>user_id</code> <strong>升序&nbsp;</strong>排序。</p>
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<p>结果格式如下所示。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例：</strong></p>
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<div class="example-block">
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<p><b>输入：</b></p>
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<p>emails 表：</p>
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<pre class="example-io">
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+----------+---------+---------------------+
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| email_id | user_id | signup_date |
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+----------+---------+---------------------+
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| 125 | 7771 | 2022-06-14 09:30:00|
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| 433 | 1052 | 2022-07-09 08:15:00|
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| 234 | 7005 | 2022-08-20 10:00:00|
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+----------+---------+---------------------+
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</pre>
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<p>texts 表：</p>
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<pre class="example-io">
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+---------+----------+--------------+---------------------+
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| text_id | email_id | signup_action| action_date |
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+---------+----------+--------------+---------------------+
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| 1 | 125 | Verified | 2022-06-15 08:30:00|
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| 2 | 433 | Not Verified | 2022-07-10 10:45:00|
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| 4 | 234 | Verified | 2022-08-21 09:30:00|
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+---------+----------+--------------+---------------------+
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</pre>
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<p><strong>输出：</strong></p>
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<pre class="example-io">
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+---------+
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| user_id |
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+---------+
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| 7005 |
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| 7771 |
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+---------+
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</pre>
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<p><strong>解释：</strong></p>
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<ul>
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<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>
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<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>
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</ul>
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</div>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一：双表关联
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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。
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<!-- tabs:start -->
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#### MySQL
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```sql
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# Write your MySQL query statement below
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SELECT user_id
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FROM
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Emails AS e
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JOIN texts AS t
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ON e.email_id = t.email_id
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AND DATEDIFF(action_date, signup_date) = 1
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AND signup_action = 'Verified'
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ORDER BY 1;
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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

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

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

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

Diff for: ‎solution/DATABASE_README.md

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| 3150 | [无效的推文 II](/solution/3100-3199/3150.Invalid%20Tweets%20II/README.md) | `数据库` | 简单 | 🔒 |
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| 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 | 🔒 |
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| 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 | 🔒 |
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| 3172 | [第二天验证](/solution/3100-3199/3172.Second%20Day%20Verification/README.md) | | 简单 | 🔒 |
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## 版权
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Diff for: ‎solution/DATABASE_README_EN.md

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| 3150 | [Invalid Tweets II](/solution/3100-3199/3150.Invalid%20Tweets%20II/README_EN.md) | `Database` | Easy | 🔒 |
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| 3156 | [Employee Task Duration and Concurrent Tasks](/solution/3100-3199/3156.Employee%20Task%20Duration%20and%20Concurrent%20Tasks/README_EN.md) | `Database` | Hard | 🔒 |
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| 3166 | [Calculate Parking Fees and Duration](/solution/3100-3199/3166.Calculate%20Parking%20Fees%20and%20Duration/README_EN.md) | `Database` | Medium | 🔒 |
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| 3172 | [Second Day Verification](/solution/3100-3199/3172.Second%20Day%20Verification/README_EN.md) | | Easy | 🔒 |
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## Copyright
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Diff for: ‎solution/README.md

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| 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 |
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| 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 |
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| 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 |
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| 3172 | [第二天验证](/solution/3100-3199/3172.Second%20Day%20Verification/README.md) | | 简单 | 🔒 |
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## 版权
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Diff for: ‎solution/README_EN.md

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| 3169 | [Count Days Without Meetings](/solution/3100-3199/3169.Count%20Days%20Without%20Meetings/README_EN.md) | | Medium | Weekly Contest 400 |
31813181
| 3170 | [Lexicographically Minimum String After Removing Stars](/solution/3100-3199/3170.Lexicographically%20Minimum%20String%20After%20Removing%20Stars/README_EN.md) | | Medium | Weekly Contest 400 |
31823182
| 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 |
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| 3172 | [Second Day Verification](/solution/3100-3199/3172.Second%20Day%20Verification/README_EN.md) | | Easy | 🔒 |
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## Copyright
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Diff for: ‎solution/contest.json

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