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Here's a well-structured `README.md` for **LeetCode 1341 - Movie Rating**, formatted for a GitHub repository:
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```md
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# 🎬 Movie Rating - LeetCode 1341
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## 📌 Problem Statement
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You are given three tables: **Movies**, **Users**, and **MovieRating**.
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Your task is to:
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1. Find the **user who has rated the greatest number of movies**.
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- In case of a tie, return the **lexicographically smaller** name.
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2. Find the **movie with the highest average rating** in **February 2020**.
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- In case of a tie, return the **lexicographically smaller** movie title.
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---
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## 📊 Table Structure
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### **Movies Table**
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| Column Name | Type |
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| ----------- | ------- |
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| movie_id | int |
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| title | varchar |
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- `movie_id` is the **primary key** (unique identifier).
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- `title` is the **name of the movie**.
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---
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### **Users Table**
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| Column Name | Type |
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| ----------- | ------- |
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| user_id | int |
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| name | varchar |
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- `user_id` is the **primary key** (unique identifier).
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- `name` is **unique** for each user.
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---
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### **MovieRating Table**
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| Column Name | Type |
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| ----------- | ---- |
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| movie_id | int |
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| user_id | int |
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| rating | int |
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| created_at | date |
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- `(movie_id, user_id)` is the **primary key** (ensuring unique user-movie ratings).
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- `created_at` represents the **review date**.
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---
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## 🔢 Goal:
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- Return a **single-column result** containing:
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1. **User name** with the most ratings.
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2. **Movie title** with the highest **average rating** in **February 2020**.
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---
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## 📊 Example 1:
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### **Input:**
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#### **Movies Table**
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| movie_id | title |
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| -------- | -------- |
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| 1 | Avengers |
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| 2 | Frozen 2 |
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| 3 | Joker |
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#### **Users Table**
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| user_id | name |
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| ------- | ------ |
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| 1 | Daniel |
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| 2 | Monica |
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| 3 | Maria |
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| 4 | James |
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#### **MovieRating Table**
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| movie_id | user_id | rating | created_at |
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| -------- | ------- | ------ | ---------- |
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| 1 | 1 | 3 | 2020年01月12日 |
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| 1 | 2 | 4 | 2020年02月11日 |
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| 1 | 3 | 2 | 2020年02月12日 |
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| 1 | 4 | 1 | 2020年01月01日 |
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| 2 | 1 | 5 | 2020年02月17日 |
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| 2 | 2 | 2 | 2020年02月01日 |
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| 2 | 3 | 2 | 2020年03月01日 |
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| 3 | 1 | 3 | 2020年02月22日 |
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| 3 | 2 | 4 | 2020年02月25日 |
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### **Output:**
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| results |
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| -------- |
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| Daniel |
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| Frozen 2 |
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### **Explanation:**
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- **Most Active User:**
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- `Daniel` and `Monica` both rated **3 movies**.
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- Since `Daniel` is **lexicographically smaller**, he is chosen.
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- **Highest Average Movie Rating in February 2020:**
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- **Frozen 2**: `(5 + 2) / 2 = 3.5`
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- **Joker**: `(3 + 4) / 2 = 3.5`
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- Since **Frozen 2** is **lexicographically smaller**, it is chosen.
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---
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## 🖥 SQL Solution
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### ✅ **Using `JOIN` + `GROUP BY` + `HAVING`**
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#### **Explanation:**
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1. **Find the most active user:**
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- Count the number of ratings per user.
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- Use `ORDER BY COUNT(*) DESC, name` to get the **user with the most ratings**, breaking ties lexicographically.
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- Limit the result to **1 user**.
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2. **Find the highest-rated movie in February 2020:**
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- Filter rows where `created_at` is **in February 2020**.
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- **Calculate the average rating per movie**.
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- Use `ORDER BY AVG(rating) DESC, title` to get the **highest-rated movie**, breaking ties lexicographically.
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- Limit the result to **1 movie**.
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```sql
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(
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SELECT name AS results
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FROM
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Users
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JOIN MovieRating USING (user_id)
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GROUP BY user_id
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ORDER BY COUNT(1) DESC, name
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LIMIT 1
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)
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UNION ALL
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(
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SELECT title
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FROM
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MovieRating
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JOIN Movies USING (movie_id)
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WHERE DATE_FORMAT(created_at, '%Y-%m') = '2020-02'
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GROUP BY movie_id
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ORDER BY AVG(rating) DESC, title
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LIMIT 1
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);
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```
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---
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## 🐍 Pandas Solution (Python)
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#### **Explanation:**
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1. **Find the user with the most ratings:**
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- Group by `user_id`, count the ratings.
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- Merge with `Users` table to get `name`.
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- Sort by **count descending**, then **lexicographically**.
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2. **Find the highest-rated movie in February 2020:**
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- Filter only `created_at` **in February 2020**.
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- Group by `movie_id` and calculate **average rating**.
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- Merge with `Movies` to get `title`.
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- Sort by **rating descending**, then **lexicographically**.
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```python
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import pandas as pd
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def movie_rating(users: pd.DataFrame, movies: pd.DataFrame, movie_rating: pd.DataFrame) -> pd.DataFrame:
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# Most active user
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user_counts = movie_rating.groupby("user_id")["rating"].count().reset_index()
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most_active_user = user_counts.merge(users, on="user_id")
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most_active_user = most_active_user.sort_values(by=["rating", "name"], ascending=[False, True]).iloc[0]["name"]
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# Highest-rated movie in February 2020
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movie_rating["created_at"] = pd.to_datetime(movie_rating["created_at"])
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feb_ratings = movie_rating[movie_rating["created_at"].dt.strftime('%Y-%m') == "2020-02"]
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avg_ratings = feb_ratings.groupby("movie_id")["rating"].mean().reset_index()
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highest_rated_movie = avg_ratings.merge(movies, on="movie_id")
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highest_rated_movie = highest_rated_movie.sort_values(by=["rating", "title"], ascending=[False, True]).iloc[0]["title"]
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return pd.DataFrame({"results": [most_active_user, highest_rated_movie]})
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```
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---
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## 📁 File Structure
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```
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📂 Movie-Rating
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│── 📜 README.md
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│── 📜 solution.sql
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│── 📜 solution_pandas.py
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│── 📜 test_cases.sql
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```
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---
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## 🔗 Useful Links
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- 📖 [LeetCode Problem](https://leetcode.com/problems/movie-rating/)
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- 📚 [SQL `GROUP BY` Clause](https://www.w3schools.com/sql/sql_groupby.asp)
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- 🐍 [Pandas GroupBy Documentation](https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.groupby.html)
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```
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### Features of this `README.md`:
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✅ **Clear problem description with tables**
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✅ **Example with step-by-step explanation**
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✅ **SQL and Pandas solutions with detailed breakdowns**
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✅ **File structure for easy organization**
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✅ **Helpful references for further learning**
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Would you like any changes or additions? 🚀

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