# Leetcodify Similar Friends Problem

## Description

LeetCode Problem 1919.

Table: Listens

``````1
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+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| user_id     | int     |
| song_id     | int     |
| day         | date    |
+-------------+---------+
There is no primary key for this table. It may contain duplicates.
Each row of this table indicates that the user user_id listened to the song song_id on the day day.
``````

Table: Friendship

``````1
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+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user1_id      | int     |
| user2_id      | int     |
+---------------+---------+
(user1_id, user2_id) is the primary key for this table.
Each row of this table indicates that the users user1_id and user2_id are friends.
Note that user1_id < user2_id.
``````

Write an SQL query to report the similar friends of Leetcodify users. A user x and user y are similar friends if:

• Users x and y are friends, and
• Users x and y listened to the same three or more different songs on the same day.

Return the result table in any order. Note that you must return the similar pairs of friends the same way they were represented in the input (i.e., always user1_id < user2_id).

The query result format is in the following example:

``````1
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Listens table:
+---------+---------+------------+
| user_id | song_id | day        |
+---------+---------+------------+
| 1       | 10      | 2021-03-15 |
| 1       | 11      | 2021-03-15 |
| 1       | 12      | 2021-03-15 |
| 2       | 10      | 2021-03-15 |
| 2       | 11      | 2021-03-15 |
| 2       | 12      | 2021-03-15 |
| 3       | 10      | 2021-03-15 |
| 3       | 11      | 2021-03-15 |
| 3       | 12      | 2021-03-15 |
| 4       | 10      | 2021-03-15 |
| 4       | 11      | 2021-03-15 |
| 4       | 13      | 2021-03-15 |
| 5       | 10      | 2021-03-16 |
| 5       | 11      | 2021-03-16 |
| 5       | 12      | 2021-03-16 |
+---------+---------+------------+

Friendship table:
+----------+----------+
| user1_id | user2_id |
+----------+----------+
| 1        | 2        |
| 2        | 4        |
| 2        | 5        |
+----------+----------+

Result table:
+----------+----------+
| user1_id | user2_id |
+----------+----------+
| 1        | 2        |
+----------+----------+

Users 1 and 2 are friends, and they listened to songs 10, 11, and 12 on the same day. They are similar friends.
Users 1 and 3 listened to songs 10, 11, and 12 on the same day, but they are not friends.
Users 2 and 4 are friends, but they did not listen to the same three different songs.
Users 2 and 5 are friends and listened to songs 10, 11, and 12, but they did not listen to them on the same day.
``````

## MySQL Solution

``````1
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select distinct a.user_id as user1_id,
b.user_id as user2_id
from Listens a
inner join Listens b
on a.user_id < b.user_id
and a.song_id = b.song_id
and a.day = b.day
and (a.user_id, b.user_id) in (select * from Friendship)
group by a.user_id, b.user_id, a.day
having count(distinct a.song_id) >= 3
``````