Formatted question description: https://leetcode.ca/all/1892.html

1892. Page Recommendations II

Level

Hard

Description

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.

Table: Likes

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| user_id     | int     |
| page_id     | int     |
+-------------+---------+
(user_id, page_id) is the primary key for this table.
Each row of this table indicates that user_id likes page_id.

You are implementing a page recommendation system for a social media website. Your system will recommend a page to user_id if the page is liked by at least one friend of user_id and is not liked by user_id.

Write an SQL query to find all the possible page recommendations for every user. Each recommendation should appear as a row in the result table with these columns:

  • user_id: The ID of the user that your system is making the recommendation to.
  • page_id: The ID of the page that will be recommended to user_id.
  • friends_likes: The number of the friends of user_id that like page_id.

Return result table in any order.

The query result format is in the following example:

Friendship table:
+----------+----------+
| user1_id | user2_id |
+----------+----------+
| 1        | 2        |
| 1        | 3        |
| 1        | 4        |
| 2        | 3        |
| 2        | 4        |
| 2        | 5        |
| 6        | 1        |
+----------+----------+
 
Likes table:
+---------+---------+
| user_id | page_id |
+---------+---------+
| 1       | 88      |
| 2       | 23      |
| 3       | 24      |
| 4       | 56      |
| 5       | 11      |
| 6       | 33      |
| 2       | 77      |
| 3       | 77      |
| 6       | 88      |
+---------+---------+

Result table:
+---------+---------+---------------+
| user_id | page_id | friends_likes |
+---------+---------+---------------+
| 1       | 77      | 2             |
| 1       | 23      | 1             |
| 1       | 24      | 1             |
| 1       | 56      | 1             |
| 1       | 33      | 1             |
| 2       | 24      | 1             |
| 2       | 56      | 1             |
| 2       | 11      | 1             |
| 2       | 88      | 1             |
| 3       | 88      | 1             |
| 3       | 23      | 1             |
| 4       | 88      | 1             |
| 4       | 77      | 1             |
| 4       | 23      | 1             |
| 5       | 77      | 1             |
| 5       | 23      | 1             |
+---------+---------+---------------+
Take user 1 as an example:
  - User 1 is friends with users 2, 3, 4, and 6.
  - Recommended pages are 23 (user 2 liked it), 24 (user 3 liked it), 56 (user 3 liked it), 33 (user 6 liked it), and 77 (user 2 and user 3 liked it).
  - Note that page 88 is not recommended because user 1 already liked it.

Another example is user 6:
  - User 6 is friends with user 1.
  - User 1 only liked page 88, but user 6 already liked it. Hence, user 6 has no recommendations.

You can recommend pages for users 2, 3, 4, and 5 using a similar process.

Solution

Select two users user1_id and user2_id from Friendship, and select the pages liked by user2_id but not liked by user1_id. Use inner join to obtain the pages recommended to user1_id.

# Write your MySQL query statement below
select f.user1_id as user_id, l.page_id, count(*) as friends_likes
    from (
        select user1_id, user2_id from Friendship union select user2_id, user1_id from Friendship
    ) as f
    inner join Likes l
    on f.user2_id = l.user_id 
    where not exists (
        select * from Likes where user_id = f.user1_id and page_id = l.page_id
    )
    group by f.user1_id, l.page_id;

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