How do you solve Friend Requests II: Who Has the Most Friends on LeetCode?

Friend Requests II: Who Has the Most Friends (LeetCode #602) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Friendship is bidirectional, meaning both users count each other as friends.

What is the brute force solution for Friend Requests II: Who Has the Most Friends?

The brute force approach for Friend Requests II: Who Has the Most Friends is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will count the number of friends for each user by checking every request in the table. This is simple but inefficient as it requires multiple passes through the data. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Friend Requests II: Who Has the Most Friends?

Friend Requests II: Who Has the Most Friends uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Friend Requests II: Who Has the Most Friends?

Always clarify the problem requirements, especially about bidirectional relationships. Think about data structures that can help you count efficiently, like hash maps. Practice writing SQL queries as they are often used in database-related problems.

What are common mistakes when solving Friend Requests II: Who Has the Most Friends?

Not considering bidirectional friendships. Overcomplicating the counting process.

#602

Friend Requests II: Who Has the Most Friends

Medium

DatabaseHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

In the optimal approach, we will use a single pass to count friends for each user using a hash map. This is efficient as it allows us to process the data in linear time.

⚙️

Algorithm

5 steps

1Step 1: Create a hash map to count friends for each user.
2Step 2: Iterate through each row in the RequestAccepted table.
3Step 3: For each requester_id and accepter_id, increment the friend count for both users in the hash map.
4Step 4: After processing all requests, find the user with the maximum friend count in a single pass.
5Step 5: Return the user ID and the maximum friend count.

solution.py18 lines

1# Full working Python code
2import sqlite3
3
4def most_friends():
5    conn = sqlite3.connect(':memory:')
6    cursor = conn.cursor()
7    cursor.execute('''CREATE TABLE RequestAccepted (requester_id INT, accepter_id INT, accept_date DATE)''')
8    cursor.executemany('INSERT INTO RequestAccepted VALUES (?, ?, ?)', [(1, 2, '2016-06-03'), (1, 3, '2016-06-08'), (2, 3, '2016-06-08'), (3, 4, '2016-06-09')])
9
10    friend_count = {}
11    for requester, accepter, _ in cursor.execute('SELECT requester_id, accepter_id FROM RequestAccepted'):
12        friend_count[requester] = friend_count.get(requester, 0) + 1
13        friend_count[accepter] = friend_count.get(accepter, 0) + 1
14
15    user_id, max_friends = max(friend_count.items(), key=lambda x: x[1])
16    return [(user_id, max_friends)]
17
18print(most_friends())

ℹ

Complexity note: The time complexity is O(n) because we only pass through the data once to count friends. The space complexity is O(n) for storing the friend counts in a hash map.

1Friendship is bidirectional, meaning both users count each other as friends.
2Using a hash map allows for efficient counting and retrieval.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.