Last updated on January 21st, 2025 at 03:43 am
Here, we see a Friend Requests II Who Has the Most Friends LeetCode Solution. This Leetcode problem is solved using MySQL and Pandas.
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Friend Requests II Who Has the Most Friends LeetCode Solution
Table of Contents
1. Problem Statement
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.
Write a solution to find the people who have the most friends and the most friends number.
The test cases are generated so that only one person has the most friends.
The result format is in the following example.
Example 1:
Input:
requester_id | accepter_id | accept_date |
1 | 2 | 2016/06/03 |
1 | 3 | 2016/06/08 |
2 | 3 | 2016/06/08 |
3 | 4 | 2016/06/09 |
Output:
id | num |
3 | 3 |
Explanation: The person with id 3 is a friend of people 1, 2, and 4, so he has three friends in total, which is the most number than any others.
2. Code Implementation in Different Languages
2.1 Friend Requests II: Who Has the Most Friends MySQL
with base as( select requester_id id from RequestAccepted union all select accepter_id id from RequestAccepted ) select id, count(*) num from base group by 1 order by 2 desc limit 1
2.2 Friend Requests II: Who Has the Most Friends Pandas
import pandas as pd def most_friends(request_accepted: pd.DataFrame) -> pd.DataFrame: req = request_accepted.requester_id.value_counts().reset_index().rename(columns ={'requester_id':'id'}) acc = request_accepted.accepter_id.value_counts().reset_index().rename(columns ={'accepter_id':'id'}) df = req.merge(acc, on ='id', how ='outer', suffixes=('_req','_acc')).fillna(0) return df.assign(num = df.count_req + df.count_acc)[['id','num']].query('num == num.max()')