Update readme.md

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LeetCode SQL 50 Solution/1341. Movie Rating/readme.md

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