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LeetCode

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Problem

Table: Listens

+-------------+---------+
| 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

+---------------+---------+
| 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 recommend friends to Leetcodify users. We recommend user x to user y if:

Users x and y are not friends, and Users x and y listened to the same three or more different songs on the same day. Note that friend recommendations are unidirectional, meaning if user x and user y should be recommended to each other, the result table should have both user x recommended to user y and user y recommended to user x. Also, note that the result table should not contain duplicates (i.e., user y should not be recommended to user x multiple times.).

Return the result table in any order.

The query result format is in the following example.

Example 1:

Input: 
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        |
+----------+----------+
Output: 
+---------+----------------+
| user_id | recommended_id |
+---------+----------------+
| 1       | 3              |
| 2       | 3              |
| 3       | 1              |
| 3       | 2              |
+---------+----------------+
Explanation: 
Users 1 and 2 listened to songs 10, 11, and 12 on the same day, but they are already friends.
Users 1 and 3 listened to songs 10, 11, and 12 on the same day. Since they are not friends, we recommend them to each other.
Users 1 and 4 did not listen to the same three songs.
Users 1 and 5 listened to songs 10, 11, and 12, but on different days.

Similarly, we can see that users 2 and 3 listened to songs 10, 11, and 12 on the same day and are not friends, so we recommend them to each other.

Code

with c as (
    select l1.user_id as uid1, l2.user_id as uid2, count(distinct l1.song_id) as ct
    from listens l1
    join listens l2
    on l1.song_id=l2.song_id and l1.day=l2.day and l1.user_id<>l2.user_id  # make sure l1.user_id != l2.user_id, we don't wanna join on user_id itself
    group by l1.user_id, l2.user_id, l1.day
    having ct>=3  # make sure the number of different songs on each day >=3
), f (uid1, uid2) as (
    select user1_id, user2_id from friendship
    union
    select user2_id, user1_id from friendship
)
select uid1 as user_id, uid2 as recommended_id
from c
where (uid1, uid2) not in (select uid1, uid2 from f)
group by uid1, uid2;