Strong Friendship - Problem

Table: Friendship

Column NameType
user1_idint
user2_idint

(user1_id, user2_id) is the primary key for this table.

Each row indicates that 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: The result table should not contain duplicates with user1_id < user2_id.

Return the result table in any order.

Table Schema

Friendship
Column Name Type Description
user1_id PK int First user in friendship (smaller ID)
user2_id PK int Second user in friendship (larger ID)
Primary Key: (user1_id, user2_id)
Note: Each row represents a friendship where user1_id < user2_id

Input & Output

Example 1 — Basic Strong Friendship
Input Table:
user1_id user2_id
1 2
1 3
2 3
1 4
2 4
3 4
1 5
2 5
3 5
4 5
Output:
user1_id user2_id
1 2
1 3
2 3
1 4
2 4
3 4
1 5
2 5
3 5
4 5
💡 Note:

Users 1 and 2 are friends with common friends: 3, 4, 5 (3 common friends). Users 1 and 3 share friends: 2, 4, 5 (3 common). Users 2 and 3 share friends: 1, 4, 5 (3 common). All three friendships qualify as strong.
Example 2 — No Strong Friendships
Input Table:
user1_id user2_id
1 2
2 3
3 4
Output:
user1_id user2_id
💡 Note:

No friendship pairs have 3 or more common friends. Each friendship has at most 1 common friend, so no strong friendships exist.
Example 3 — Mixed Scenario
Input Table:
user1_id user2_id
1 7
1 2
1 3
2 7
3 7
2 4
3 4
Output:
user1_id user2_id
💡 Note:

Users 1 and 7 have common friends: 2, 3 (only 2 common friends, but if there were more relationships, it would be 3+). In this case, only friendship 1-7 has exactly the required number of mutual connections.

Constraints

  • 1 ≤ user1_id < user2_id ≤ 500
  • All pairs in the table represent valid friendships
  • The result should not contain duplicates

Visualization

Tap to expand
Strong Friendship Problem INPUT Friendship Table 1 2 3 4 5 6 user1_id | user2_id 1,2 | 1,3 | 2,3 | 3,4 | 3,5 3,6 | 4,5 | 5,6 | 2,6... ALGORITHM STEPS 1 Self-Join Friendship Find all pairs (x,y) who are already friends 2 Find Common Friends Join to find mutual connections between x,y 3 Count Common Friends GROUP BY user pairs COUNT(*) mutual friends 4 Filter Strong (>=3) HAVING COUNT(*) >= 3 WHERE user1 < user2 Common Friends: A B C 3+ = Strong! FINAL RESULT Strong Friendship Found 1 2 STRONG Common Friends: 3 5 6 Output Table user1_id | user2_id 1 | 2 OK - 3+ Common Key Insight: Use self-join on Friendship table to find pairs who are direct friends, then join again to count mutual friends. Filter with HAVING COUNT(*) >= 3 and ensure user1_id < user2_id to avoid duplicates. Time Complexity: O(n^2) where n is number of friendships | SQL: Self-join + GROUP BY + HAVING TutorialsPoint - Strong Friendship | Optimal Solution

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