Skip to content

3058. Friends With No Mutual Friends πŸ”’

Description

Table: Friends

+-------------+------+
| Column Name | Type |
+-------------+------+
| user_id1    | int  |
| user_id2    | int  |
+-------------+------+
(user_id1, user_id2) is the primary key (combination of columns with unique values) for this table.
Each row contains user id1, user id2, both of whom are friends with each other.

Write a solution to find all pairs of users who are friends with each other and have no mutual friends.

Return the result table ordered by user_id1, user_id2 in ascending order.

The result format is in the following example.

 

Example 1:

Input: 
Friends table:
+----------+----------+
| user_id1 | user_id2 | 
+----------+----------+
| 1        | 2        | 
| 2        | 3        | 
| 2        | 4        | 
| 1        | 5        | 
| 6        | 7        | 
| 3        | 4        | 
| 2        | 5        | 
| 8        | 9        | 
+----------+----------+
Output: 
+----------+----------+
| user_id1 | user_id2 | 
+----------+----------+
| 6        | 7        | 
| 8        | 9        | 
+----------+----------+
Explanation: 
- Users 1 and 2 are friends with each other, but they share a mutual friend with user ID 5, so this pair is not included.
- Users 2 and 3 are friends, they both share a mutual friend with user ID 4, resulting in exclusion, similarly for users 2 and 4 who share a mutual friend with user ID 3, hence not included.
- Users 1 and 5 are friends with each other, but they share a mutual friend with user ID 2, so this pair is not included.
- Users 6 and 7, as well as users 8 and 9, are friends with each other, and they don't have any mutual friends, hence included.
- Users 3 and 4 are friends with each other, but their mutual connection with user ID 2 means they are not included, similarly for users 2 and 5 are friends but are excluded due to their mutual connection with user ID 1.
Output table is ordered by user_id1 in ascending order.

Solutions

Solution 1: Subquery

First, we list all the friend relationships and record them in table T. Then we find the pairs of friends who do not have common friends.

Next, we can use a subquery to find pairs of friends who do not have common friends, i.e., this pair of friends does not belong to any other person's friends.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
# Write your MySQL query statement below
WITH
    T AS (
        SELECT user_id1, user_id2 FROM Friends
        UNION ALL
        SELECT user_id2, user_id1 FROM Friends
    )
SELECT user_id1, user_id2
FROM Friends
WHERE
    (user_id1, user_id2) NOT IN (
        SELECT t1.user_id1, t2.user_id1
        FROM
            T AS t1
            JOIN T AS t2 ON t1.user_id2 = t2.user_id2
    )
ORDER BY 1, 2;

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
import pandas as pd


def friends_with_no_mutual_friends(friends: pd.DataFrame) -> pd.DataFrame:
    cp = friends.copy()
    t = cp[["user_id1", "user_id2"]].copy()
    t = pd.concat(
        [
            t,
            cp[["user_id2", "user_id1"]].rename(
                columns={"user_id2": "user_id1", "user_id1": "user_id2"}
            ),
        ]
    )
    merged = t.merge(t, left_on="user_id2", right_on="user_id2")
    ans = cp[
        ~cp.apply(
            lambda x: (x["user_id1"], x["user_id2"])
            in zip(merged["user_id1_x"], merged["user_id1_y"]),
            axis=1,
        )
    ]
    return ans.sort_values(by=["user_id1", "user_id2"])

Comments