README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1900-1999/1949.Strong%20Friendship/README_EN.md	Database

1949. Strong Friendship 🔒

中文文档

Description

Table: Friendship

+-------------+------+
| Column Name | Type |
+-------------+------+
| user1_id    | int  |
| user2_id    | int  |
+-------------+------+
(user1_id, user2_id) is the primary key (combination of columns with unique values) 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 a solution 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 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).

Solutions

Solution 1

MySQL

# Write your MySQL query statement below
WITH
    t AS (
        SELECT
            *
        FROM Friendship
        UNION ALL
        SELECT
            user2_id,
            user1_id
        FROM Friendship
    )
SELECT
    t1.user1_id,
    t1.user2_id,
    COUNT(1) AS common_friend
FROM
    t AS t1
    JOIN t AS t2 ON t1.user2_id = t2.user1_id
    JOIN t AS t3 ON t1.user1_id = t3.user1_id
WHERE t3.user2_id = t2.user2_id AND t1.user1_id < t1.user2_id
GROUP BY t1.user1_id, t1.user2_id
HAVING COUNT(1) >= 3;