# All the Pairs With the Maximum Number of Common Followers Problem

## Description

LeetCode Problem 1951.

Table: Relations

``````1
2
3
4
5
6
7
8
+-------------+------+
| Column Name | Type |
+-------------+------+
| user_id     | int  |
| follower_id | int  |
+-------------+------+
(user_id, follower_id) is the primary key for this table.
Each row of this table indicates that the user with ID follower_id is following the user with ID user_id.
``````

Write an SQL query to find all the pairs of users with the maximum number of common followers. In other words, if the maximum number of common followers between any two users is maxCommon, then you have to return all pairs of users that have maxCommon common followers.

The result table should contain the pairs user1_id and user2_id where user1_id < user2_id.

Return the result table in any order.

The query result format is in the following example:

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
Relations table:
+---------+-------------+
| user_id | follower_id |
+---------+-------------+
| 1       | 3           |
| 2       | 3           |
| 7       | 3           |
| 1       | 4           |
| 2       | 4           |
| 7       | 4           |
| 1       | 5           |
| 2       | 6           |
| 7       | 5           |
+---------+-------------+

Result table:
+----------+----------+
| user1_id | user2_id |
+----------+----------+
| 1        | 7        |
+----------+----------+

Users 1 and 2 have 2 common followers (3 and 4).
Users 1 and 7 have 3 common followers (3, 4, and 5).
Users 2 and 7 have 2 common followers (3 and 4).
Since the maximum number of common followers between any two users is 3, we return all pairs of users with 3 common followers, which is only the pair (1, 7). We return the pair as (1, 7), not as (7, 1).
Note that we do not have any information about the users that follow users 3, 4, and 5, so we consider them to have 0 followers.
``````

## MySQL Solution

``````1
2
3
4
5
6
7
8
select user1_id, user2_id
from (
select r1.user_id as user1_id, r2.user_id as user2_id, dense_rank() over (order by count(*) desc) rk
from Relations r1
join Relations r2
on r1.user_id < r2.user_id and r1.follower_id = r2.follower_id
group by r1.user_id, r2.user_id) temp
where rk=1
``````