# 3059. Find All Unique Email Domains

## Description

Table: Emails

+-------------+---------+
\| Column Name \| Type    \|
+-------------+---------+
\| id          \| int     \|
\| email       \| varchar \|
+-------------+---------+
id is the primary key (column with unique values) for this table.
Each row of this table contains an email. The emails will not contain uppercase letters.


Write a solution to find all unique email domains and count the number of individuals associated with each domain. Consider only those domains that end with .com.

Return the result table orderd by email domains in ascending order.

The result format is in the following example.

Example 1:

Input:
Emails table:
+-----+-----------------------+
\| id  \| email                 \|
+-----+-----------------------+
\| 336 \| hwkiy@test.edu        \|
\| 449 \| vrzmwyum@yahoo.com    \|
\| 95  \| tof@test.edu          \|
\| 320 \| jxhbagkpm@example.org \|
\| 411 \| zxcf@outlook.com      \|
+----+------------------------+
Output:
+--------------+-------+
\| email_domain \| count \|
+--------------+-------+
\| outlook.com  \| 2     \|
\| yahoo.com    \| 1     \|
+--------------+-------+
Explanation:
- The valid domains ending with ".com" are only "outlook.com" and "yahoo.com", with respective counts of 2 and 1.
Output table is ordered by email_domains in ascending order.


## Solutions

### Solution 1: Using SUBSTRING_INDEX Function + Grouping Statistics

First, we filter out all emails ending with .com, then use the SUBSTRING_INDEX function to extract the domain name of the email. Finally, we use GROUP BY to count the number of each domain.

• import pandas as pd

def find_unique_email_domains(emails: pd.DataFrame) -> pd.DataFrame:
emails["email_domain"] = emails["email"].str.split("@").str[-1]
emails = emails[emails["email"].str.contains(".com")]
return (
emails.groupby("email_domain")
.size()
.reset_index(name="count")
.sort_values(by="email_domain")
)


• # Write your MySQL query statement below
SELECT SUBSTRING_INDEX(email, '@', -1) AS email_domain, COUNT(1) AS count
FROM Emails
WHERE email LIKE '%.com'
GROUP BY 1
ORDER BY 1;