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  • LeetCode SQL 50 Solution/1075. Project Employees I

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

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