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  • LeetCode SQL 50 Solution/1141. User Activity for the Past 30 Days I

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Here’s a well-structured `README.md` file for **LeetCode 1141 - User Activity for the Past 30 Days I** with SQL and Pandas solutions:
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```md
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# 📊 User Activity for the Past 30 Days I - LeetCode 1141
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## 📌 Problem Statement
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You are given the **Activity** table that records user activities on a social media website.
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### Activity Table
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| Column Name | Type |
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| ------------- | ---- |
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| user_id | int |
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| session_id | int |
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| activity_date | date |
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| activity_type | enum |
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- The `activity_type` column is an ENUM of **('open_session', 'end_session', 'scroll_down', 'send_message')**.
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- Each session belongs to exactly **one user**.
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- The table **may have duplicate rows**.
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### Task:
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Find the **daily active user count** for a period of **30 days ending 2019年07月27日 inclusively**.
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- A user is considered **active on a given day** if they made at least **one activity**.
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- Ignore days with **zero active users**.
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---
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## 📊 Example 1:
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### Input:
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**Activity Table**
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| user_id | session_id | activity_date | activity_type |
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| ------- | ---------- | ------------- | ------------- |
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| 1 | 1 | 2019年07月20日 | open_session |
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| 1 | 1 | 2019年07月20日 | scroll_down |
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| 1 | 1 | 2019年07月20日 | end_session |
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| 2 | 4 | 2019年07月20日 | open_session |
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| 2 | 4 | 2019年07月21日 | send_message |
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| 2 | 4 | 2019年07月21日 | end_session |
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| 3 | 2 | 2019年07月21日 | open_session |
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| 3 | 2 | 2019年07月21日 | send_message |
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| 3 | 2 | 2019年07月21日 | end_session |
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| 4 | 3 | 2019年06月25日 | open_session |
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| 4 | 3 | 2019年06月25日 | end_session |
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### Output:
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| day | active_users |
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| ---------- | ------------ |
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| 2019年07月20日 | 2 |
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| 2019年07月21日 | 2 |
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### Explanation:
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- **2019年07月20日**: Users **1 and 2** were active.
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- **2019年07月21日**: Users **2 and 3** were active.
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- **Days with zero active users are ignored**.
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---
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## 🖥 SQL Solutions
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### 1️⃣ Standard MySQL Solution
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#### Explanation:
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- **Filter records** for the last **30 days** (ending on `2019年07月27日`).
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- Use `COUNT(DISTINCT user_id)` to count **unique active users per day**.
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- Ignore **days with zero active users**.
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```sql
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SELECT
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activity_date AS day,
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COUNT(DISTINCT user_id) AS active_users
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FROM
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Activity
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WHERE
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DATEDIFF('2019年07月27日', activity_date) < 30
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AND DATEDIFF('2019年07月27日', activity_date) >= 0
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GROUP BY activity_date;
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```
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---
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### 2️⃣ Alternative Solution Using `BETWEEN`
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#### Explanation:
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- This solution filters the date range using `BETWEEN` instead of `DATEDIFF`.
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```sql
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SELECT
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activity_date AS day,
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COUNT(DISTINCT user_id) AS active_users
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FROM
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Activity
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WHERE
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activity_date BETWEEN DATE_SUB('2019年07月27日', INTERVAL 29 DAY) AND '2019年07月27日'
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GROUP BY activity_date;
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```
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---
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## 🐍 Pandas Solution (Python)
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#### Explanation:
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- Filter activity records for the **last 30 days**.
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- **Group by `activity_date`** and count **unique `user_id`s**.
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- **Ignore days with zero active users**.
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```python
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import pandas as pd
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def daily_active_users(activity: pd.DataFrame) -> pd.DataFrame:
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# Filter data within the last 30 days (ending on '2019年07月27日')
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filtered = activity[(activity["activity_date"] >= "2019年06月28日") & (activity["activity_date"] <= "2019年07月27日")]
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# Group by day and count unique users
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result = filtered.groupby("activity_date")["user_id"].nunique().reset_index()
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# Rename columns
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result.columns = ["day", "active_users"]
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return result
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```
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---
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## 📁 File Structure
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```
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📂 User-Activity-Past-30-Days
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│── 📜 README.md
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│── 📜 solution.sql
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│── 📜 solution_between.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/user-activity-for-the-past-30-days-i/)
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- 📚 [SQL Date Functions](https://www.w3schools.com/sql/sql_dates.asp)
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- 🐍 [Pandas Documentation](https://pandas.pydata.org/docs/)
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```
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This README includes:
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- **Clear problem statement**
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- **Input and output tables**
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- **Multiple SQL solutions with explanations**
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- **Pandas solution in Python**
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- **File structure for a clean GitHub repo**
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- **Useful links**
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Let me know if you need any changes! 🚀

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