1949. Strong Friendship
Problem
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.
A friendship between a pair of friends x and y is strong if x and y have at least three common friends.
Write an SQL query to find all the strong friendships.
Note that the result table should not contain duplicates with user1_id < user2_id.
Return the result table in any order.
The query result format is in the following example.
Example 1:
Input:
Friendship table:
+----------+----------+
| user1_id | user2_id |
+----------+----------+
| 1 | 2 |
| 1 | 3 |
| 2 | 3 |
| 1 | 4 |
| 2 | 4 |
| 1 | 5 |
| 2 | 5 |
| 1 | 7 |
| 3 | 7 |
| 1 | 6 |
| 3 | 6 |
| 2 | 6 |
+----------+----------+
Output:
+----------+----------+---------------+
| user1_id | user2_id | common_friend |
+----------+----------+---------------+
| 1 | 2 | 4 |
| 1 | 3 | 3 |
+----------+----------+---------------+
Explanation:
Users 1 and 2 have 4 common friends (3, 4, 5, and 6).
Users 1 and 3 have 3 common friends (2, 6, and 7).
We did not include the friendship of users 2 and 3 because they only have two common friends (1 and 6).
Code
with f as (
select user1_id, user2_id
from Friendship
union
select user2_id user1_id, user1_id user2_id
from Friendship
)
select a.user1_id, a.user2_id, count(c.user2_id) common_friend
from Friendship a
join f b
on a.user1_id = b.user1_id # u1 friends
join f c
on a.user2_id = c.user1_id # u2 friends
and b.user2_id = c.user2_id # u1 u2 comman friends
group by a.user1_id, a.user2_id
having count(c.user2_id) >= 3