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feat: add sql solution to lc problem: No.3451 #4057

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1 change: 1 addition & 0 deletions .prettierignore
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Expand Up @@ -26,3 +26,4 @@ node_modules/
/solution/3100-3199/3150.Invalid Tweets II/Solution.sql
/solution/3100-3199/3198.Find Cities in Each State/Solution.sql
/solution/3300-3399/3328.Find Cities in Each State II/Solution.sql
/solution/3400-3499/3451.Find Invalid IP Addresses/Solution.sql
162 changes: 162 additions & 0 deletions solution/3400-3499/3451.Find Invalid IP Addresses/README.md
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---
comments: true
difficulty: 困难
edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3451.Find%20Invalid%20IP%20Addresses/README.md
tags:
- 数据库
---

<!-- problem:start -->

# [3451. Find Invalid IP Addresses](https://leetcode.cn/problems/find-invalid-ip-addresses)

[English Version](/solution/3400-3499/3451.Find%20Invalid%20IP%20Addresses/README_EN.md)

## 题目描述

<!-- description:start -->

<p>Table: <code> logs</code></p>

<pre>
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| log_id | int |
| ip | varchar |
| status_code | int |
+-------------+---------+
log_id is the unique key for this table.
Each row contains server access log information including IP address and HTTP status code.
</pre>

<p>Write a solution to find <strong>invalid IP addresses</strong>. An IPv4 address is invalid if it meets any of these conditions:</p>

<ul>
<li>Contains numbers <strong>greater than</strong> <code>255</code> in any octet</li>
<li>Has <strong>leading zeros</strong> in any octet (like <code>01.02.03.04</code>)</li>
<li>Has <strong>less or more</strong> than <code>4</code> octets</li>
</ul>

<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>

<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>logs table:</p>

<pre class="example-io">
+--------+---------------+-------------+
| log_id | ip | status_code |
+--------+---------------+-------------+
| 1 | 192.168.1.1 | 200 |
| 2 | 256.1.2.3 | 404 |
| 3 | 192.168.001.1 | 200 |
| 4 | 192.168.1.1 | 200 |
| 5 | 192.168.1 | 500 |
| 6 | 256.1.2.3 | 404 |
| 7 | 192.168.001.1 | 200 |
+--------+---------------+-------------+
</pre>

<p><strong>Output:</strong></p>

<pre class="example-io">
+---------------+--------------+
| ip | invalid_count|
+---------------+--------------+
| 256.1.2.3 | 2 |
| 192.168.001.1 | 2 |
| 192.168.1 | 1 |
+---------------+--------------+
</pre>

<p><strong>Explanation:</strong></p>

<ul>
<li>256.1.2.3&nbsp;is invalid because 256 &gt; 255</li>
<li>192.168.001.1&nbsp;is invalid because of leading zeros</li>
<li>192.168.1&nbsp;is invalid because it has only 3 octets</li>
</ul>

<p>The output table is ordered by invalid_count, ip in descending order respectively.</p>
</div>

<!-- description:end -->

## 解法

<!-- solution:start -->

### 方法一：模拟

我们可以根据题意，判断 IP 地址是否不合法，判断的条件有：

1. IP 地址中的 `.` 的个数不等于 $3$；
2. IP 地址中的某个 octet 以 `0` 开头；
3. IP 地址中的某个 octet 大于 $255$。

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。

<!-- tabs:start -->

#### MySQL

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

#### Pandas

```python
import pandas as pd


def find_invalid_ips(logs: pd.DataFrame) -> pd.DataFrame:
def is_valid_ip(ip: str) -> bool:
octets = ip.split(".")
if len(octets) != 4:
return False
for octet in octets:
if not octet.isdigit():
return False
value = int(octet)
if not 0 <= value <= 255 or octet != str(value):
return False
return True

logs["is_valid"] = logs["ip"].apply(is_valid_ip)
invalid_ips = logs[~logs["is_valid"]]
invalid_count = invalid_ips["ip"].value_counts().reset_index()
invalid_count.columns = ["ip", "invalid_count"]
result = invalid_count.sort_values(
by=["invalid_count", "ip"], ascending=[False, False]
)
return result
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
162 changes: 162 additions & 0 deletions solution/3400-3499/3451.Find Invalid IP Addresses/README_EN.md
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---
comments: true
difficulty: Hard
edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3451.Find%20Invalid%20IP%20Addresses/README_EN.md
tags:
- Database
---

<!-- problem:start -->

# [3451. Find Invalid IP Addresses](https://leetcode.com/problems/find-invalid-ip-addresses)

[中文文档](/solution/3400-3499/3451.Find%20Invalid%20IP%20Addresses/README.md)

## Description

<!-- description:start -->

<p>Table: <code> logs</code></p>

<pre>
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| log_id | int |
| ip | varchar |
| status_code | int |
+-------------+---------+
log_id is the unique key for this table.
Each row contains server access log information including IP address and HTTP status code.
</pre>

<p>Write a solution to find <strong>invalid IP addresses</strong>. An IPv4 address is invalid if it meets any of these conditions:</p>

<ul>
<li>Contains numbers <strong>greater than</strong> <code>255</code> in any octet</li>
<li>Has <strong>leading zeros</strong> in any octet (like <code>01.02.03.04</code>)</li>
<li>Has <strong>less or more</strong> than <code>4</code> octets</li>
</ul>

<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>

<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>logs table:</p>

<pre class="example-io">
+--------+---------------+-------------+
| log_id | ip | status_code |
+--------+---------------+-------------+
| 1 | 192.168.1.1 | 200 |
| 2 | 256.1.2.3 | 404 |
| 3 | 192.168.001.1 | 200 |
| 4 | 192.168.1.1 | 200 |
| 5 | 192.168.1 | 500 |
| 6 | 256.1.2.3 | 404 |
| 7 | 192.168.001.1 | 200 |
+--------+---------------+-------------+
</pre>

<p><strong>Output:</strong></p>

<pre class="example-io">
+---------------+--------------+
| ip | invalid_count|
+---------------+--------------+
| 256.1.2.3 | 2 |
| 192.168.001.1 | 2 |
| 192.168.1 | 1 |
+---------------+--------------+
</pre>

<p><strong>Explanation:</strong></p>

<ul>
<li>256.1.2.3&nbsp;is invalid because 256 &gt; 255</li>
<li>192.168.001.1&nbsp;is invalid because of leading zeros</li>
<li>192.168.1&nbsp;is invalid because it has only 3 octets</li>
</ul>

<p>The output table is ordered by invalid_count, ip in descending order respectively.</p>
</div>

<!-- description:end -->

## Solutions

<!-- solution:start -->

### Solution 1: Simulation

We can determine if an IP address is invalid based on the following conditions:

1. The number of `.` in the IP address is not equal to $3$;
2. Any octet in the IP address starts with `0`;
3. Any octet in the IP address is greater than $255$.

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.

<!-- tabs:start -->

#### MySQL

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

#### Pandas

```python
import pandas as pd


def find_invalid_ips(logs: pd.DataFrame) -> pd.DataFrame:
def is_valid_ip(ip: str) -> bool:
octets = ip.split(".")
if len(octets) != 4:
return False
for octet in octets:
if not octet.isdigit():
return False
value = int(octet)
if not 0 <= value <= 255 or octet != str(value):
return False
return True

logs["is_valid"] = logs["ip"].apply(is_valid_ip)
invalid_ips = logs[~logs["is_valid"]]
invalid_count = invalid_ips["ip"].value_counts().reset_index()
invalid_count.columns = ["ip", "invalid_count"]
result = invalid_count.sort_values(
by=["invalid_count", "ip"], ascending=[False, False]
)
return result
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
24 changes: 24 additions & 0 deletions solution/3400-3499/3451.Find Invalid IP Addresses/Solution.py
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import pandas as pd


def find_invalid_ips(logs: pd.DataFrame) -> pd.DataFrame:
def is_valid_ip(ip: str) -> bool:
octets = ip.split(".")
if len(octets) != 4:
return False
for octet in octets:
if not octet.isdigit():
return False
value = int(octet)
if not 0 <= value <= 255 or octet != str(value):
return False
return True

logs["is_valid"] = logs["ip"].apply(is_valid_ip)
invalid_ips = logs[~logs["is_valid"]]
invalid_count = invalid_ips["ip"].value_counts().reset_index()
invalid_count.columns = ["ip", "invalid_count"]
result = invalid_count.sort_values(
by=["invalid_count", "ip"], ascending=[False, False]
)
return result
19 changes: 19 additions & 0 deletions solution/3400-3499/3451.Find Invalid IP Addresses/Solution.sql
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SELECT
ip,
COUNT(*) AS invalid_count
FROM logs
WHERE
LENGTH(ip) - LENGTH(REPLACE(ip, '.', '')) != 3

OR SUBSTRING_INDEX(ip, '.', 1) REGEXP '^0[0-9]'
OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 2), '.', -1) REGEXP '^0[0-9]'
OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 3), '.', -1) REGEXP '^0[0-9]'
OR SUBSTRING_INDEX(ip, '.', -1) REGEXP '^0[0-9]'

OR SUBSTRING_INDEX(ip, '.', 1) > 255
OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 2), '.', -1) > 255
OR SUBSTRING_INDEX(SUBSTRING_INDEX(ip, '.', 3), '.', -1) > 255
OR SUBSTRING_INDEX(ip, '.', -1) > 255

GROUP BY 1
ORDER BY 2 DESC, 1 DESC;
1 change: 1 addition & 0 deletions solution/DATABASE_README.md
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Expand Up @@ -309,6 +309,7 @@
| 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 | 🔒 |
| 3421 | [查找进步的学生](/solution/3400-3499/3421.Find%20Students%20Who%20Improved/README.md) | `数据库` | 中等 | |
| 3436 | [查找合法邮箱](/solution/3400-3499/3436.Find%20Valid%20Emails/README.md) | `数据库` | 简单 | |
| 3451 | [Find Invalid IP Addresses](/solution/3400-3499/3451.Find%20Invalid%20IP%20Addresses/README.md) | | 困难 | |

## 版权

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1 change: 1 addition & 0 deletions solution/DATABASE_README_EN.md
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Expand Up @@ -307,6 +307,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
| 3415 | [Find Products with Three Consecutive Digits](/solution/3400-3499/3415.Find%20Products%20with%20Three%20Consecutive%20Digits/README_EN.md) | `Database` | Easy | 🔒 |
| 3421 | [Find Students Who Improved](/solution/3400-3499/3421.Find%20Students%20Who%20Improved/README_EN.md) | `Database` | Medium | |
| 3436 | [Find Valid Emails](/solution/3400-3499/3436.Find%20Valid%20Emails/README_EN.md) | `Database` | Easy | |
| 3451 | [Find Invalid IP Addresses](/solution/3400-3499/3451.Find%20Invalid%20IP%20Addresses/README_EN.md) | | Hard | |

## Copyright

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1 change: 1 addition & 0 deletions solution/README.md
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Expand Up @@ -3461,6 +3461,7 @@
| 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 |
| 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 |
| 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 | 🔒 |
| 3451 | [Find Invalid IP Addresses](/solution/3400-3499/3451.Find%20Invalid%20IP%20Addresses/README.md) | | 困难 | |

## 版权

Expand Down
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