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feat: add sql solution to lc problem: No.3451 (#4057)
No.3451.Find Invalid IP Addresses
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‎.prettierignore

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/solution/3100-3199/3150.Invalid Tweets II/Solution.sql
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/solution/3100-3199/3198.Find Cities in Each State/Solution.sql
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/solution/3300-3399/3328.Find Cities in Each State II/Solution.sql
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/solution/3400-3499/3451.Find Invalid IP Addresses/Solution.sql

‎solution/3400-3499/3451.Find Invalid IP Addresses/README.md

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---
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comments: true
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difficulty: 困难
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3451.Find%20Invalid%20IP%20Addresses/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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# [3451. Find Invalid IP Addresses](https://leetcode.cn/problems/find-invalid-ip-addresses)
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[English Version](/solution/3400-3499/3451.Find%20Invalid%20IP%20Addresses/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>Table: <code> logs</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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| log_id | int |
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| ip | varchar |
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| status_code | int |
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+-------------+---------+
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log_id is the unique key for this table.
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Each row contains server access log information including IP address and HTTP status code.
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</pre>
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<p>Write a solution to find <strong>invalid IP addresses</strong>. An IPv4 address is invalid if it meets any of these conditions:</p>
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<ul>
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<li>Contains numbers <strong>greater than</strong> <code>255</code> in any octet</li>
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<li>Has <strong>leading zeros</strong> in any octet (like <code>01.02.03.04</code>)</li>
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<li>Has <strong>less or more</strong> than <code>4</code> octets</li>
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</ul>
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<p>Return <em>the result table </em><em>ordered by</em> <code>invalid_count</code>,&nbsp;<code>ip</code>&nbsp;<em>in <strong>descending</strong> order respectively</em>.&nbsp;</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>logs table:</p>
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<pre class="example-io">
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+--------+---------------+-------------+
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| log_id | ip | status_code |
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+--------+---------------+-------------+
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| 1 | 192.168.1.1 | 200 |
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| 2 | 256.1.2.3 | 404 |
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| 3 | 192.168.001.1 | 200 |
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| 4 | 192.168.1.1 | 200 |
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| 5 | 192.168.1 | 500 |
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| 6 | 256.1.2.3 | 404 |
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| 7 | 192.168.001.1 | 200 |
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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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| ip | invalid_count|
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+---------------+--------------+
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| 256.1.2.3 | 2 |
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| 192.168.001.1 | 2 |
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| 192.168.1 | 1 |
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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>256.1.2.3&nbsp;is invalid because 256 &gt; 255</li>
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<li>192.168.001.1&nbsp;is invalid because of leading zeros</li>
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<li>192.168.1&nbsp;is invalid because it has only 3 octets</li>
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</ul>
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<p>The output table is ordered by invalid_count, ip in descending order respectively.</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：模拟
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我们可以根据题意，判断 IP 地址是否不合法，判断的条件有：
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1. IP 地址中的 `.` 的个数不等于 $3$；
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2. IP 地址中的某个 octet 以 `0` 开头；
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3. IP 地址中的某个 octet 大于 $255$。
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然后我们将不合法的 IP 地址进行分组，并统计每个不合法的 IP 地址的个数 `invalid_count`，最后按照 `invalid_count` 和 `ip` 降序排序。
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<!-- tabs:start -->
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#### MySQL
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```sql
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SELECT
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ip,
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COUNT(*) AS invalid_count
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FROM logs
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WHERE
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LENGTH(ip) - LENGTH(REPLACE(ip, '.', '')) != 3
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OR SUBSTRING_INDEX(ip, '.', 1) REGEXP '^0[0-9]'
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OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 2), '.', -1) REGEXP '^0[0-9]'
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OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 3), '.', -1) REGEXP '^0[0-9]'
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OR SUBSTRING_INDEX(ip, '.', -1) REGEXP '^0[0-9]'
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OR SUBSTRING_INDEX(ip, '.', 1) > 255
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OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 2), '.', -1) > 255
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OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 3), '.', -1) > 255
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OR SUBSTRING_INDEX(ip, '.', -1) > 255
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GROUP BY 1
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ORDER BY 2 DESC, 1 DESC;
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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_invalid_ips(logs: pd.DataFrame) -> pd.DataFrame:
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def is_valid_ip(ip: str) -> bool:
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octets = ip.split(".")
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if len(octets) != 4:
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return False
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for octet in octets:
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if not octet.isdigit():
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return False
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value = int(octet)
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if not 0 <= value <= 255 or octet != str(value):
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return False
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return True
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logs["is_valid"] = logs["ip"].apply(is_valid_ip)
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invalid_ips = logs[~logs["is_valid"]]
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invalid_count = invalid_ips["ip"].value_counts().reset_index()
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invalid_count.columns = ["ip", "invalid_count"]
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result = invalid_count.sort_values(
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by=["invalid_count", "ip"], ascending=[False, False]
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)
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return result
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/3400-3499/3451.Find Invalid IP Addresses/README_EN.md

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---
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comments: true
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difficulty: Hard
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3451.Find%20Invalid%20IP%20Addresses/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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# [3451. Find Invalid IP Addresses](https://leetcode.com/problems/find-invalid-ip-addresses)
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[中文文档](/solution/3400-3499/3451.Find%20Invalid%20IP%20Addresses/README.md)
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## Description
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<!-- description:start -->
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<p>Table: <code> logs</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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| log_id | int |
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| ip | varchar |
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| status_code | int |
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+-------------+---------+
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log_id is the unique key for this table.
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Each row contains server access log information including IP address and HTTP status code.
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</pre>
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<p>Write a solution to find <strong>invalid IP addresses</strong>. An IPv4 address is invalid if it meets any of these conditions:</p>
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<ul>
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<li>Contains numbers <strong>greater than</strong> <code>255</code> in any octet</li>
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<li>Has <strong>leading zeros</strong> in any octet (like <code>01.02.03.04</code>)</li>
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<li>Has <strong>less or more</strong> than <code>4</code> octets</li>
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</ul>
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<p>Return <em>the result table </em><em>ordered by</em> <code>invalid_count</code>,&nbsp;<code>ip</code>&nbsp;<em>in <strong>descending</strong> order respectively</em>.&nbsp;</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>logs table:</p>
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<pre class="example-io">
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+--------+---------------+-------------+
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| log_id | ip | status_code |
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+--------+---------------+-------------+
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| 1 | 192.168.1.1 | 200 |
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| 2 | 256.1.2.3 | 404 |
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| 3 | 192.168.001.1 | 200 |
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| 4 | 192.168.1.1 | 200 |
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| 5 | 192.168.1 | 500 |
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| 6 | 256.1.2.3 | 404 |
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| 7 | 192.168.001.1 | 200 |
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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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| ip | invalid_count|
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+---------------+--------------+
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| 256.1.2.3 | 2 |
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| 192.168.001.1 | 2 |
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| 192.168.1 | 1 |
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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>256.1.2.3&nbsp;is invalid because 256 &gt; 255</li>
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<li>192.168.001.1&nbsp;is invalid because of leading zeros</li>
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<li>192.168.1&nbsp;is invalid because it has only 3 octets</li>
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</ul>
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<p>The output table is ordered by invalid_count, ip in descending order respectively.</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: Simulation
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We can determine if an IP address is invalid based on the following conditions:
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1. The number of `.` in the IP address is not equal to $3$;
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2. Any octet in the IP address starts with `0`;
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3. Any octet in the IP address is greater than $255$.
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Then we group the invalid IP addresses and count the occurrences of each invalid IP address `invalid_count`, and finally sort by `invalid_count` and `ip` in descending order.
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<!-- tabs:start -->
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#### MySQL
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```sql
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SELECT
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ip,
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COUNT(*) AS invalid_count
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FROM logs
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WHERE
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LENGTH(ip) - LENGTH(REPLACE(ip, '.', '')) != 3
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OR SUBSTRING_INDEX(ip, '.', 1) REGEXP '^0[0-9]'
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OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 2), '.', -1) REGEXP '^0[0-9]'
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OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 3), '.', -1) REGEXP '^0[0-9]'
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OR SUBSTRING_INDEX(ip, '.', -1) REGEXP '^0[0-9]'
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OR SUBSTRING_INDEX(ip, '.', 1) > 255
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OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 2), '.', -1) > 255
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OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 3), '.', -1) > 255
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OR SUBSTRING_INDEX(ip, '.', -1) > 255
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GROUP BY 1
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ORDER BY 2 DESC, 1 DESC;
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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_invalid_ips(logs: pd.DataFrame) -> pd.DataFrame:
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def is_valid_ip(ip: str) -> bool:
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octets = ip.split(".")
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if len(octets) != 4:
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return False
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for octet in octets:
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if not octet.isdigit():
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return False
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value = int(octet)
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if not 0 <= value <= 255 or octet != str(value):
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return False
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return True
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logs["is_valid"] = logs["ip"].apply(is_valid_ip)
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invalid_ips = logs[~logs["is_valid"]]
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invalid_count = invalid_ips["ip"].value_counts().reset_index()
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invalid_count.columns = ["ip", "invalid_count"]
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result = invalid_count.sort_values(
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by=["invalid_count", "ip"], ascending=[False, False]
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)
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return result
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/3400-3499/3451.Find Invalid IP Addresses/Solution.py

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import pandas as pd
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def find_invalid_ips(logs: pd.DataFrame) -> pd.DataFrame:
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def is_valid_ip(ip: str) -> bool:
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octets = ip.split(".")
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if len(octets) != 4:
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return False
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for octet in octets:
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if not octet.isdigit():
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return False
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value = int(octet)
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if not 0 <= value <= 255 or octet != str(value):
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return False
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return True
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logs["is_valid"] = logs["ip"].apply(is_valid_ip)
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invalid_ips = logs[~logs["is_valid"]]
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invalid_count = invalid_ips["ip"].value_counts().reset_index()
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invalid_count.columns = ["ip", "invalid_count"]
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result = invalid_count.sort_values(
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by=["invalid_count", "ip"], ascending=[False, False]
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)
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return result

‎solution/3400-3499/3451.Find Invalid IP Addresses/Solution.sql

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SELECT
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ip,
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COUNT(*) AS invalid_count
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FROM logs
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WHERE
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LENGTH(ip) - LENGTH(REPLACE(ip, '.', '')) != 3
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OR SUBSTRING_INDEX(ip, '.', 1) REGEXP '^0[0-9]'
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OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 2), '.', -1) REGEXP '^0[0-9]'
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OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 3), '.', -1) REGEXP '^0[0-9]'
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OR SUBSTRING_INDEX(ip, '.', -1) REGEXP '^0[0-9]'
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OR SUBSTRING_INDEX(ip, '.', 1) > 255
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OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 2), '.', -1) > 255
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OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 3), '.', -1) > 255
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OR SUBSTRING_INDEX(ip, '.', -1) > 255
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GROUP BY 1
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ORDER BY 2 DESC, 1 DESC;

‎solution/DATABASE_README.md

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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 | [查找合法邮箱](/solution/3400-3499/3436.Find%20Valid%20Emails/README.md) | `数据库` | 简单 | |
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| 3451 | [Find Invalid IP Addresses](/solution/3400-3499/3451.Find%20Invalid%20IP%20Addresses/README.md) | | 困难 | |
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## 版权
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‎solution/DATABASE_README_EN.md

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@@ -307,6 +307,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
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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 | |
309309
| 3436 | [Find Valid Emails](/solution/3400-3499/3436.Find%20Valid%20Emails/README_EN.md) | `Database` | Easy | |
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| 3451 | [Find Invalid IP Addresses](/solution/3400-3499/3451.Find%20Invalid%20IP%20Addresses/README_EN.md) | | Hard | |
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## Copyright
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‎solution/README.md

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34613461
| 3448 | [统计可以被最后一个数位整除的子字符串数目](/solution/3400-3499/3448.Count%20Substrings%20Divisible%20By%20Last%20Digit/README.md) | | 困难 | 第 436 场周赛 |
34623462
| 3449 | [最大化游戏分数的最小值](/solution/3400-3499/3449.Maximize%20the%20Minimum%20Game%20Score/README.md) | | 困难 | 第 436 场周赛 |
34633463
| 3450 | [一张长椅的上最多学生](/solution/3400-3499/3450.Maximum%20Students%20on%20a%20Single%20Bench/README.md) | | 简单 | 🔒 |
3464+
| 3451 | [Find Invalid IP Addresses](/solution/3400-3499/3451.Find%20Invalid%20IP%20Addresses/README.md) | | 困难 | |
34643465

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## 版权
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