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Exercise 24:1917.Leetcodify Friends Recommendations

1.Description

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.

2.Create Table and insert into values

create table if not exists leetcode.ex_1917_listens
(user_id string , song_id string, day date ) stored as orc ;
​
INSERT INTO table leetcode.ex_1917_listens VALUES
('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') ;
​
create table if not exists leetcode.ex_1917_friendship
(user1_id string , user2_id string ) stored as orc ;
​
INSERT INTO table leetcode.ex_1917_friendship VALUES
('1','2') ;
In order to prevent the stability of big data clusters. Similar non-congruent joins (non-inner joins) are forbidden, and SemanticException Cartesian products are disabled.
IF we want non-congruent joins, Add two settings before the query:
set hive.strict.checks.cartesian.product=flase;
set hive.mapred.mode=nonstrict;
​
Check this link for more detail.
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