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LeetCode SQL 50 Solution/1075. Project Employees I/readme.md

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+
+# 🏆 Project Employees I - LeetCode 1075
+
+## 📌 Problem Statement
+You are given two tables: **Project** and **Employee**.
+
+### Project Table
+| Column Name | Type |
+| ----------- | ---- |
+| project_id  | int  |
+| employee_id | int  |
+
+- `(project_id, employee_id)` is the primary key of this table.
+- `employee_id` is a foreign key referencing the `Employee` table.
+
+### Employee Table
+| Column Name      | Type    |
+| ---------------- | ------- |
+| employee_id      | int     |
+| name             | varchar |
+| experience_years | int     |
+
+- `employee_id` is the primary key.
+- `experience_years` is guaranteed to be **NOT NULL**.
+
+The task is to **return the average experience years of all employees for each project, rounded to 2 decimal places**.
+
+---
+
+## 📊 Example 1:
+### Input:
+**Project Table**
+| project_id | employee_id |
+| ---------- | ----------- |
+| 1          | 1           |
+| 1          | 2           |
+| 1          | 3           |
+| 2          | 1           |
+| 2          | 4           |
+
+**Employee Table**
+| employee_id | name   | experience_years |
+| ----------- | ------ | ---------------- |
+| 1           | Khaled | 3                |
+| 2           | Ali    | 2                |
+| 3           | John   | 1                |
+| 4           | Doe    | 2                |
+
+### Output:
+| project_id | average_years |
+| ---------- | ------------- |
+| 1          | 2.00          |
+| 2          | 2.50          |
+
+### Explanation:
+- **Project 1:** `(3 + 2 + 1) / 3 = 2.00`
+- **Project 2:** `(3 + 2) / 2 = 2.50`
+
+---
+
+## 🖥 SQL Solutions
+
+### 1️⃣ Standard MySQL Solution
+#### Explanation:
+- We **JOIN** the `Project` and `Employee` tables using `employee_id`.
+- We **calculate the average** of `experience_years` for each `project_id`.
+- We **round** the result to **two decimal places**.
+
+```sql
+SELECT project_id, ROUND(AVG(experience_years), 2) AS average_years
+FROM project AS p
+LEFT JOIN employee AS e
+ON p.employee_id = e.employee_id
+GROUP BY project_id;
+```
+
+---
+
+### 2️⃣ Window Function (SQL) Solution
+#### Explanation:
+- Using **window functions**, we calculate the `AVG(experience_years)` over a **partitioned** dataset.
+
+```sql
+SELECT DISTINCT project_id, 
+       ROUND(AVG(experience_years) OVER (PARTITION BY project_id), 2) AS average_years
+FROM project AS p
+JOIN employee AS e 
+ON p.employee_id = e.employee_id;
+```
+
+---
+
+## 🐍 Pandas Solution (Python)
+#### Explanation:
+- We read both tables into Pandas **DataFrames**.
+- We merge the tables on `employee_id`.
+- We group by `project_id` and compute the mean.
+- We round the output to 2 decimal places.
+
+```python
+import pandas as pd
+
+def project_average_experience(project: pd.DataFrame, employee: pd.DataFrame) -> pd.DataFrame:
+    df = project.merge(employee, on="employee_id")
+    result = df.groupby("project_id")["experience_years"].mean().round(2).reset_index()
+    result.columns = ["project_id", "average_years"]
+    return result
+```
+
+---
+
+## 📁 File Structure
+```
+📂 Project-Employees-I
+│── 📜 README.md
+│── 📜 solution.sql
+│── 📜 solution_window.sql
+│── 📜 solution_pandas.py
+│── 📜 test_cases.sql
+```
+
+---
+
+## 🔗 Useful Links
+- 📖 [LeetCode Problem](https://leetcode.com/problems/project-employees-i/)
+- 📚 [SQL Joins Explanation](https://www.w3schools.com/sql/sql_join.asp)
+- 🐍 [Pandas Documentation](https://pandas.pydata.org/docs/)
+
+---
+
+## Let me know if you need any modifications! 🚀

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

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