# $1264. 页面推荐
# https://leetcode-cn.com/problems/page-recommendations/

# SQL架构
Create table If Not Exists Friendship (user1_id int, user2_id int);
Create table If Not Exists Likes (user_id int, page_id int);
Truncate table Friendship;
insert into Friendship (user1_id, user2_id) values ('1', '2');
insert into Friendship (user1_id, user2_id) values ('1', '3');
insert into Friendship (user1_id, user2_id) values ('1', '4');
insert into Friendship (user1_id, user2_id) values ('2', '3');
insert into Friendship (user1_id, user2_id) values ('2', '4');
insert into Friendship (user1_id, user2_id) values ('2', '5');
insert into Friendship (user1_id, user2_id) values ('6', '1');
Truncate table Likes;
insert into Likes (user_id, page_id) values ('1', '88');
insert into Likes (user_id, page_id) values ('2', '23');
insert into Likes (user_id, page_id) values ('3', '24');
insert into Likes (user_id, page_id) values ('4', '56');
insert into Likes (user_id, page_id) values ('5', '11');
insert into Likes (user_id, page_id) values ('6', '33');
insert into Likes (user_id, page_id) values ('2', '77');
insert into Likes (user_id, page_id) values ('3', '77');
insert into Likes (user_id, page_id) values ('6', '88');

# Write your MySQL query statement below
select
    distinct page_id as recommended_page
from
    Likes
where
    user_id in (
        select
            user1_id as user_id
        from
            Friendship
        where
            user2_id = 1
        union
        all
        select
            user2_id as user_id
        from
            Friendship
        where
            user1_id = 1
    )
    and page_id not in (
        select
            page_id
        from
            Likes
        where
            user_id = 1
    );

# clean-up
drop table Friendship;
drop table Likes;