Merge pull request #80 from iamAntimPal/Branch-1

Branch 1

Author: iamAntimPal

LeetCode SQL 50 Solution/602. Friend Requests II/602. Friend Requests II.py

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+import pandas as pd
+
+def most_friends(requests: pd.DataFrame) -> pd.DataFrame:
+    # Create a DataFrame that contains all friend relationships in both directions
+    friend_df = pd.concat([
+        requests[['requester_id', 'accepter_id']].rename(columns={'requester_id': 'id', 'accepter_id': 'friend'}),
+        requests[['accepter_id', 'requester_id']].rename(columns={'accepter_id': 'id', 'requester_id': 'friend'})
+    ])
+    
+    # Count number of friends for each user
+    friend_counts = friend_df.groupby('id').size().reset_index(name='num')
+    
+    # Get the user with the most friends
+    max_friends = friend_counts.loc[friend_counts['num'].idxmax()]
+    
+    return pd.DataFrame({'id': [max_friends['id']], 'num': [max_friends['num']]})

LeetCode SQL 50 Solution/602. Friend Requests II/readme.md

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+# **602. Friend Requests II: Who Has the Most Friends**
+
+## **Problem Statement**
+You are given a table `RequestAccepted` that records friend request acceptances between users.
+
+### **RequestAccepted Table**
+```
++--------------+-------------+-------------+
+| Column Name  | Type        |
++--------------+-------------+
+| requester_id | int         |
+| accepter_id  | int         |
+| accept_date  | date        |
++--------------+-------------+
+```
+- **(requester_id, accepter_id)** is the **primary key**.
+- Each row indicates the user who sent a friend request (`requester_id`), the user who accepted it (`accepter_id`), and the date when the request was accepted.
+
+### **Task:**
+Find the person who has the **most friends** along with the number of friends they have.  
+*Note:* The test cases are generated so that only one person has the most friends.
+
+---
+
+## **Example 1:**
+
+### **Input:**
+#### **RequestAccepted Table**
+```
++--------------+-------------+-------------+
+| 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:**
+- User with `id = 3` is friends with users 1, 2, and 4, making a total of **3 friends**—the highest among all users.
+
+---
+
+## **Solution Approaches**
+
+### **SQL Solution (Using UNION ALL)**
+```sql
+WITH T AS (
+    SELECT requester_id, accepter_id FROM RequestAccepted
+    UNION ALL
+    SELECT accepter_id, requester_id FROM RequestAccepted
+)
+SELECT requester_id AS id, COUNT(*) AS num
+FROM T
+GROUP BY requester_id
+ORDER BY num DESC
+LIMIT 1;
+```
+**Explanation:**
+- The CTE `T` creates a complete friendship list by combining both directions of the friend relationship.
+- The outer query groups by `requester_id` (which now represents a user) and counts the number of occurrences (i.e., friends).
+- The result is ordered by the friend count in descending order and limited to one row, returning the user with the most friends.
+
+---
+
+### **Pandas Solution**
+```python
+import pandas as pd
+
+def most_friends(requests: pd.DataFrame) -> pd.DataFrame:
+    # Create a DataFrame that contains all friend relationships in both directions
+    friend_df = pd.concat([
+        requests[['requester_id', 'accepter_id']].rename(columns={'requester_id': 'id', 'accepter_id': 'friend'}),
+        requests[['accepter_id', 'requester_id']].rename(columns={'accepter_id': 'id', 'requester_id': 'friend'})
+    ])
+    
+    # Count number of friends for each user
+    friend_counts = friend_df.groupby('id').size().reset_index(name='num')
+    
+    # Get the user with the most friends
+    max_friends = friend_counts.loc[friend_counts['num'].idxmax()]
+    
+    return pd.DataFrame({'id': [max_friends['id']], 'num': [max_friends['num']]})
+
+# Example usage:
+# requests_df = pd.read_csv('request_accepted.csv')
+# print(most_friends(requests_df))
+```
+**Explanation:**
+- The solution concatenates two DataFrames to consider friend relationships in both directions.
+- It then groups by user `id` and counts the number of friends.
+- The user with the maximum friend count is selected and returned.
+
+---
+
+## **File Structure**
+```
+LeetCode602/
+├── problem_statement.md       # Contains the problem description and constraints.
+├── sql_solution.sql           # Contains the SQL solution using UNION ALL.
+├── pandas_solution.py         # Contains the Pandas solution for Python users.
+├── README.md                  # Overview of the problem and available solutions.
+```
+
+---
+
+## **Useful Links**
+- [LeetCode Problem 602](https://leetcode.com/problems/friend-requests-ii-who-has-the-most-friends/)
+- [SQL UNION ALL Documentation](https://www.w3schools.com/sql/sql_union.asp)
+- [Pandas concat Documentation](https://pandas.pydata.org/docs/reference/api/pandas.concat.html)
+- [Pandas groupby Documentation](https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.groupby.html)