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feat: add solutions to lc problem: No.3436 (doocs#4004)
No.3436.Find Valid Emails
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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/3400-3499/3436.Find%20Valid%20Emails/README.md
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tags:
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- 数据库
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---
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<!-- problem:start -->
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# [3436. Find Valid Emails](https://leetcode.cn/problems/find-valid-emails)
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[English Version](/solution/3400-3499/3436.Find%20Valid%20Emails/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>Table: <code>Users</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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| user_id | int |
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| email | varchar |
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+-----------------+---------+
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(user_id) is the unique key for this table.
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Each row contains a user&#39;s unique ID and email address.
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</pre>
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<p>Write a solution to find all the <strong>valid email addresses</strong>. A valid email address meets the following criteria:</p>
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<ul>
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<li>It contains exactly one <code>@</code> symbol.</li>
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<li>The part before the <code>@</code> symbol contains only <strong>alphanumeric</strong> characters and <strong>underscores</strong>.</li>
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<li>The part after the <code>@</code> symbol contains a domain name <strong>that starts with a letter</strong> and ends with <code>.com</code>.</li>
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</ul>
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<p>Return<em> the result table ordered by</em> <code>user_id</code> <em>in</em> <strong>ascending </strong><em>order</em>.</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>Users table:</p>
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<pre class="example-io">
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+---------+---------------------+
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| user_id | email |
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+---------+---------------------+
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| 1 | alice@example.com |
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| 2 | bob_at_example.com |
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| 3 | charlie@example.net |
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| 4 | david@domain.com |
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| 5 | eve@invalid |
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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 | email |
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+---------+-------------------+
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| 1 | alice@example.com |
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| 4 | david@domain.com |
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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><strong>alice@example.com</strong> is valid because it contains one <code>@</code>, alice&nbsp;is alphanumeric, and example.com&nbsp;starts with a letter and ends with .com.</li>
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<li><strong>bob_at_example.com</strong> is invalid because it contains an underscore instead of an <code>@</code>.</li>
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<li><strong>charlie@example.net</strong> is invalid because the domain does not end with <code>.com</code>.</li>
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<li><strong>david@domain.com</strong> is valid because it meets all criteria.</li>
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<li><strong>eve@invalid</strong> is invalid because the domain does not end with <code>.com</code>.</li>
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</ul>
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<p>Result table is ordered by user_id in ascending order.</p>
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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:正则表达式
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我们可以使用正则表达式,通过 `REGEXP` 来匹配符合条件的邮箱地址。
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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, email
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FROM Users
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WHERE email REGEXP '^[A-Za-z0-9_]+@[A-Za-z][A-Za-z0-9]*\\.com$'
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ORDER BY 1;
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```
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#### Pandas
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```python
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import pandas as pd
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def find_valid_emails(users: pd.DataFrame) -> pd.DataFrame:
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email_pattern = r"^[A-Za-z0-9_]+@[A-Za-z][A-Za-z0-9]*\.com$"
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valid_emails = users[users["email"].str.match(email_pattern)]
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valid_emails = valid_emails.sort_values(by="user_id")
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return valid_emails
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->
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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/3400-3499/3436.Find%20Valid%20Emails/README_EN.md
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tags:
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- Database
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---
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<!-- problem:start -->
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# [3436. Find Valid Emails](https://leetcode.com/problems/find-valid-emails)
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[中文文档](/solution/3400-3499/3436.Find%20Valid%20Emails/README.md)
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## Description
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<!-- description:start -->
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<p>Table: <code>Users</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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| user_id | int |
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| email | varchar |
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+-----------------+---------+
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(user_id) is the unique key for this table.
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Each row contains a user&#39;s unique ID and email address.
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</pre>
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<p>Write a solution to find all the <strong>valid email addresses</strong>. A valid email address meets the following criteria:</p>
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<ul>
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<li>It contains exactly one <code>@</code> symbol.</li>
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<li>The part before the <code>@</code> symbol contains only <strong>alphanumeric</strong> characters and <strong>underscores</strong>.</li>
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<li>The part after the <code>@</code> symbol contains a domain name <strong>that starts with a letter</strong> and ends with <code>.com</code>.</li>
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</ul>
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<p>Return<em> the result table ordered by</em> <code>user_id</code> <em>in</em> <strong>ascending </strong><em>order</em>.</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>Users table:</p>
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<pre class="example-io">
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+---------+---------------------+
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| user_id | email |
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+---------+---------------------+
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| 1 | alice@example.com |
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| 2 | bob_at_example.com |
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| 3 | charlie@example.net |
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| 4 | david@domain.com |
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| 5 | eve@invalid |
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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 | email |
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+---------+-------------------+
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| 1 | alice@example.com |
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| 4 | david@domain.com |
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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><strong>alice@example.com</strong> is valid because it contains one <code>@</code>, alice&nbsp;is alphanumeric, and example.com&nbsp;starts with a letter and ends with .com.</li>
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<li><strong>bob_at_example.com</strong> is invalid because it contains an underscore instead of an <code>@</code>.</li>
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<li><strong>charlie@example.net</strong> is invalid because the domain does not end with <code>.com</code>.</li>
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<li><strong>david@domain.com</strong> is valid because it meets all criteria.</li>
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<li><strong>eve@invalid</strong> is invalid because the domain does not end with <code>.com</code>.</li>
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</ul>
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<p>Result table is ordered by user_id in ascending order.</p>
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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: Regular Expression
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We can use a regular expression with `REGEXP` to match valid email addresses.
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The time complexity is $O(n),ドル and the space complexity is $O(1)$. Here, $n$ is the length of the input string.
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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, email
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FROM Users
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WHERE email REGEXP '^[A-Za-z0-9_]+@[A-Za-z][A-Za-z0-9]*\\.com$'
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ORDER BY 1;
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```
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#### Pandas
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```python
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import pandas as pd
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def find_valid_emails(users: pd.DataFrame) -> pd.DataFrame:
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email_pattern = r"^[A-Za-z0-9_]+@[A-Za-z][A-Za-z0-9]*\.com$"
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valid_emails = users[users["email"].str.match(email_pattern)]
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valid_emails = valid_emails.sort_values(by="user_id")
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return valid_emails
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->
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import pandas as pd
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def find_valid_emails(users: pd.DataFrame) -> pd.DataFrame:
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email_pattern = r"^[A-Za-z0-9_]+@[A-Za-z][A-Za-z0-9]*\.com$"
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valid_emails = users[users["email"].str.match(email_pattern)]
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valid_emails = valid_emails.sort_values(by="user_id")
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return valid_emails
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# Write your MySQL query statement below
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SELECT user_id, email
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FROM Users
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WHERE email REGEXP '^[A-Za-z0-9_]+@[A-Za-z][A-Za-z0-9]*\\.com$'
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ORDER BY 1;

‎solution/DATABASE_README.md‎

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| 3401 | [Find Circular Gift Exchange Chains](/solution/3400-3499/3401.Find%20Circular%20Gift%20Exchange%20Chains/README.md) | `数据库` | 困难 | 🔒 |
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| 3415 | [查找具有三个连续数字的产品](/solution/3400-3499/3415.Find%20Products%20with%20Three%20Consecutive%20Digits/README.md) | `数据库` | 简单 | 🔒 |
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| 3421 | [查找进步的学生](/solution/3400-3499/3421.Find%20Students%20Who%20Improved/README.md) | `数据库` | 中等 | |
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| 3436 | [Find Valid Emails](/solution/3400-3499/3436.Find%20Valid%20Emails/README.md) | | 简单 | |
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## 版权
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‎solution/DATABASE_README_EN.md‎

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| 3401 | [Find Circular Gift Exchange Chains](/solution/3400-3499/3401.Find%20Circular%20Gift%20Exchange%20Chains/README_EN.md) | `Database` | Hard | 🔒 |
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| 3415 | [Find Products with Three Consecutive Digits](/solution/3400-3499/3415.Find%20Products%20with%20Three%20Consecutive%20Digits/README_EN.md) | `Database` | Easy | 🔒 |
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| 3421 | [Find Students Who Improved](/solution/3400-3499/3421.Find%20Students%20Who%20Improved/README_EN.md) | `Database` | Medium | |
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| 3436 | [Find Valid Emails](/solution/3400-3499/3436.Find%20Valid%20Emails/README_EN.md) | | Easy | |
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## Copyright
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‎solution/README.md‎

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| 3433 | [统计用户被提及情况](/solution/3400-3499/3433.Count%20Mentions%20Per%20User/README.md) | `数组`,`数学`,`排序`,`模拟` | 中等 | 第 434 场周赛 |
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| 3434 | [子数组操作后的最大频率](/solution/3400-3499/3434.Maximum%20Frequency%20After%20Subarray%20Operation/README.md) | `贪心`,`数组`,`哈希表`,`动态规划`,`前缀和` | 中等 | 第 434 场周赛 |
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| 3435 | [最短公共超序列的字母出现频率](/solution/3400-3499/3435.Frequencies%20of%20Shortest%20Supersequences/README.md) | `位运算`,`图`,`拓扑排序`,`数组`,`字符串`,`枚举` | 困难 | 第 434 场周赛 |
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| 3436 | [Find Valid Emails](/solution/3400-3499/3436.Find%20Valid%20Emails/README.md) | | 简单 | |
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## 版权
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‎solution/README_EN.md‎

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| 3433 | [Count Mentions Per User](/solution/3400-3499/3433.Count%20Mentions%20Per%20User/README_EN.md) | `Array`,`Math`,`Sorting`,`Simulation` | Medium | Weekly Contest 434 |
34453445
| 3434 | [Maximum Frequency After Subarray Operation](/solution/3400-3499/3434.Maximum%20Frequency%20After%20Subarray%20Operation/README_EN.md) | `Greedy`,`Array`,`Hash Table`,`Dynamic Programming`,`Prefix Sum` | Medium | Weekly Contest 434 |
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| 3435 | [Frequencies of Shortest Supersequences](/solution/3400-3499/3435.Frequencies%20of%20Shortest%20Supersequences/README_EN.md) | `Bit Manipulation`,`Graph`,`Topological Sort`,`Array`,`String`,`Enumeration` | Hard | Weekly Contest 434 |
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| 3436 | [Find Valid Emails](/solution/3400-3499/3436.Find%20Valid%20Emails/README_EN.md) | | Easy | |
34473448

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## Copyright
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