Solution Explanation

This problem requires finding all pairs of users with the maximum number of common followers from the Relations table. The solution uses a SQL query with a common table expression (CTE) to efficiently achieve this.

Approach

1. Self-Join: The query starts by joining the Relations table with itself using JOIN Relations AS r2 ON r1.follower_id = r2.follower_id AND r1.user_id < r2.user_id. This self-join finds pairs of users (r1.user_id, r2.user_id) who share at least one common follower. The condition r1.user_id < r2.user_id ensures that each pair is counted only once, avoiding duplicates like (1,7) and (7,1).

2. Counting Common Followers: GROUP BY r1.user_id, r2.user_id groups the results by user pairs. COUNT(1) counts the number of common followers for each pair.

3. Ranking: RANK() OVER (ORDER BY COUNT(1) DESC) AS rk assigns a rank to each pair based on the number of common followers. Pairs with the most common followers receive rank 1. RANK() handles ties, assigning the same rank to pairs with equal numbers of common followers.

4. Filtering for Maximum: Finally, SELECT user1_id, user2_id FROM t WHERE rk = 1 selects only the user pairs with rank 1 (those having the maximum number of common followers).

Time Complexity Analysis

The time complexity of this SQL query is dominated by the self-join operation. In the worst case, the number of pairs to compare is O(n²), where n is the number of distinct user IDs. The GROUP BY, COUNT(), and RANK() operations have a time complexity proportional to the number of grouped rows, which is at most O(n²). Therefore, the overall time complexity is O(n²). The space complexity is also O(n²) in the worst case due to the potential size of the intermediate result sets.

Code in Other Languages (Conceptual)

While the problem is inherently SQL-based, we can outline how a similar approach might be implemented in other languages conceptually. The core idea remains the same: finding all pairs, counting common followers for each pair, and then identifying pairs with the maximum count.

Python (Illustrative - requires a suitable database library like SQLAlchemy):

# Conceptual Python code - requires a database interaction library
# This code snippet is illustrative and would need adaptation based on database library used.
 
# ... (Database connection and fetching data from Relations table) ...
 
user_data = {user_id: set(follower_ids) for user_id, follower_ids in relations_data}
 
max_common = 0
max_pairs = []
 
for user1_id, followers1 in user_data.items():
    for user2_id in user_data.keys():
        if user1_id < user2_id:
          common = len(followers1.intersection(user_data[user2_id]))
          if common > max_common:
              max_common = common
              max_pairs = [(user1_id, user2_id)]
          elif common == max_common:
              max_pairs.append((user1_id, user2_id))
 
 
print(max_pairs)  # Output: List of user pairs with maximum common followers
 
# ... (Database connection closure) ...

The Python code provides a conceptual outline. The actual implementation would heavily rely on the chosen database interaction library for efficient data retrieval and processing. The time complexity would still be roughly O(n²) due to the nested loops iterating through user pairs.

Similar conceptual approaches could be developed in other languages, but SQL remains the most natural and efficient choice for this database-centric problem.