Leetcode #602: Friend Requests II: Who Has the Most Friends
In this guide, we solve Leetcode #602 Friend Requests II: Who Has the Most Friends in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Table: RequestAccepted +----------------+---------+ | Column Name | Type | +----------------+---------+ | requester_id | int | | accepter_id | int | | accept_date | date | +----------------+---------+ (requester_id, accepter_id) is the primary key (combination of columns with unique values) for this table. This table contains the ID of the user who sent the request, the ID of the user who received the request, and the date when the request was accepted.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Database
Intuition
The task is relational in nature, which maps cleanly to DataFrame operations in Python.
By treating tables as DataFrames, joins and group-bys become concise and readable.
Approach
Load the inputs as DataFrames and apply the appropriate merge, filter, or group-by.
Select or rename the columns to match the required output.
Steps:
- Load inputs as DataFrames.
- Apply merge/groupby/filter operations.
- Select the output columns.
Example
+----------------+---------+
| Column Name | Type |
+----------------+---------+
| requester_id | int |
| accepter_id | int |
| accept_date | date |
+----------------+---------+
(requester_id, accepter_id) is the primary key (combination of columns with unique values) for this table.
This table contains the ID of the user who sent the request, the ID of the user who received the request, and the date when the request was accepted.
Python Solution
import pandas as pd
def most_friends(request_accepted: pd.DataFrame) -> pd.DataFrame:
ids = pd.concat([request_accepted['requester_id'], request_accepted['accepter_id']], ignore_index=True)
counts = ids.value_counts()
max_cnt = counts.max()
person = counts[counts == max_cnt].index.min()
return pd.DataFrame({'id': [int(person)], 'num': [int(max_cnt)]})
Complexity
The time complexity is O(n log n) (typical). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.