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  • LeetCode SQL 50 Solution/1731. The Number of Employees Which Report to Each Employee

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Here’s a well-structured `README.md` for **LeetCode 1731 - The Number of Employees Which Report to Each Employee**, formatted for a GitHub repository:
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```md
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# 👥 The Number of Employees Which Report to Each Employee - LeetCode 1731
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## 📌 Problem Statement
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You are given a table **Employees** that contains the following columns:
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- `employee_id`: The unique ID of the employee.
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- `name`: The name of the employee.
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- `reports_to`: The `employee_id` of the manager the employee reports to (can be `NULL` if the employee does not report to anyone).
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- `age`: The age of the employee.
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A manager is defined as an employee who has **at least 1 direct report**.
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Your task is to report:
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- The **IDs and names of all managers**.
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- The **number of employees** who report **directly** to them.
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- The **average age** of their direct reports, rounded to the nearest integer.
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Return the result **ordered by `employee_id`** in ascending order.
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---
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## 📊 Table Structure
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### **Employees 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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| reports_to | int |
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| age | int |
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- `employee_id` is the **primary key**.
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- `reports_to` may be `NULL` for employees who do not report to anyone.
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---
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## 📊 Example 1:
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### **Input:**
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| employee_id | name | reports_to | age |
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| ----------- | ------- | ---------- | --- |
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| 9 | Hercy | NULL | 43 |
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| 6 | Alice | 9 | 41 |
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| 4 | Bob | 9 | 36 |
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| 2 | Winston | NULL | 37 |
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### **Output:**
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| employee_id | name | reports_count | average_age |
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| ----------- | ----- | ------------- | ----------- |
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| 9 | Hercy | 2 | 39 |
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### **Explanation:**
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- **Hercy** (employee_id = 9) is a manager with two direct reports: **Alice** (age 41) and **Bob** (age 36).
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- The average age of these reports is (41 + 36) / 2 = 38.5, which is rounded to **39**.
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---
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## 🖥 SQL Solution
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### ✅ **Approach:**
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1. Use a **self-join** on the **Employees** table where the employee’s `reports_to` matches the manager’s `employee_id`.
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2. Count the number of direct reports for each manager.
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3. Compute the average age of the direct reports and round the result to the nearest integer.
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4. Group by the manager’s `employee_id` and order the results by `employee_id`.
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```sql
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SELECT
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Manager.employee_id,
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Manager.name,
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COUNT(Employee.employee_id) AS reports_count,
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ROUND(AVG(Employee.age)) AS average_age
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FROM Employees AS Manager
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INNER JOIN Employees AS Employee
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ON Employee.reports_to = Manager.employee_id
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GROUP BY Manager.employee_id
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ORDER BY Manager.employee_id;
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```
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---
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## 🐍 Python (Pandas) Solution
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### ✅ **Approach:**
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1. Filter the DataFrame to create a join between managers and their direct reports.
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2. Group by the manager’s `employee_id` and compute:
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- The count of direct reports.
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- The average age of the reports, then round the average.
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3. Merge the results with the original manager information.
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4. Sort the result by `employee_id`.
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```python
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import pandas as pd
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def employees_reporting(employees: pd.DataFrame) -> pd.DataFrame:
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# Merge the table with itself: one for managers and one for employees reporting to them.
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merged = employees.merge(
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employees,
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left_on='employee_id',
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right_on='reports_to',
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suffixes=('_manager', '_report')
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)
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# Group by manager's employee_id and name, then compute the count and average age of reports.
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result = merged.groupby(['employee_id_manager', 'name_manager']).agg(
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reports_count=('employee_id_report', 'count'),
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average_age=('age_report', lambda x: round(x.mean()))
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).reset_index()
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# Rename columns to match expected output.
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result.rename(columns={
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'employee_id_manager': 'employee_id',
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'name_manager': 'name'
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}, inplace=True)
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# Sort by employee_id in ascending order.
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result = result.sort_values('employee_id').reset_index(drop=True)
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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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📂 Employees-Reporting
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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/the-number-of-employees-which-report-to-each-employee/)
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- 🔍 [MySQL GROUP BY Documentation](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 table structure**
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✅ **Example with detailed explanation**
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✅ **SQL solution with breakdown**
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✅ **Python (Pandas) solution with code and explanation**
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✅ **Organized repository structure**
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✅ **Helpful links for further learning**
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Would you like any further modifications?

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