Merge pull request #76 from iamAntimPal/Branch-1

Branch 1

Author: iamAntimPal

LeetCode SQL 50 Solution/570. Managers with at Least 5 Direct Reports/570. Managers with at Least 5 Direct Reports.py

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+import pandas as pd
+
+def managers_with_five_reports(employee: pd.DataFrame) -> None:
+    # Count direct reports for each manager
+    report_counts = employee.groupby('managerId').size().reset_index(name='report_count')
+    
+    # Identify managerIds with at least five direct reports
+    valid_managers = report_counts[report_counts['report_count'] >= 5]['managerId']
+    
+    # Filter the Employee table to get manager names
+    # Note: Since managerId can be null, we ignore them during merge.
+    result = employee[employee['id'].isin(valid_managers)][['name']]
+    
+    # Modify the original DataFrame in place if required.
+    employee = result
+    print(result)

LeetCode SQL 50 Solution/570. Managers with at Least 5 Direct Reports/readme.md

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+# **570. Managers with at Least 5 Direct Reports**
+
+## **Problem Statement**
+You are given a table `Employee` that holds all employee records, including their managers. Every employee has an `id`, a `name`, a `department`, and a `managerId`.
+
+### **Employee Table**
+```
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| id          | int     |
+| name        | varchar |
+| department  | varchar |
+| managerId   | int     |
++-------------+---------+
+```
+- `id` is the **primary key**.
+- `managerId` is a **foreign key** that references the `id` of a manager in the same table.
+- If `managerId` is `null`, then the employee does not have a manager.
+- No employee will be the manager of themselves.
+
+### **Task:**
+Write a solution to find all managers (i.e., employees who appear as a managerId in the table) with **at least five direct reports**.
+
+The result table should display the manager’s name in any order.
+
+---
+
+## **Example 1:**
+
+### **Input:**
+#### **Employee Table**
+```
++-----+-------+------------+-----------+
+| id  | name  | department | managerId |
++-----+-------+------------+-----------+
+| 101 | John  | A          | null      |
+| 102 | Dan   | A          | 101       |
+| 103 | James | A          | 101       |
+| 104 | Amy   | A          | 101       |
+| 105 | Anne  | A          | 101       |
+| 106 | Ron   | B          | 101       |
++-----+-------+------------+-----------+
+```
+
+### **Output:**
+```
++------+
+| name |
++------+
+| John |
++------+
+```
+
+### **Explanation:**
+- **John** (id = 101) is the only manager who has **5 direct reports** (ids 102, 103, 104, 105, and 106).
+
+---
+
+## **Solution Approaches**
+
+### **SQL Solution (Using Subquery)**
+```sql
+SELECT name
+FROM Employee
+WHERE id IN (
+    SELECT managerId
+    FROM Employee
+    GROUP BY managerId
+    HAVING COUNT(*) >= 5
+);
+```
+**Explanation:**
+- The subquery groups the `Employee` table by `managerId` and counts the number of direct reports.
+- Only managers with a count of **5 or more** are selected.
+- The outer query then retrieves the names of those managers.
+
+---
+
+### **SQL Solution (Using JOIN and Window Functions)**
+```sql
+SELECT name
+FROM Employee
+JOIN (
+    SELECT managerId
+    FROM Employee
+    GROUP BY managerId
+    HAVING COUNT(*) >= 5
+) AS managers
+ON Employee.id = managers.managerId;
+```
+**Explanation:**
+- The inner query identifies all `managerId`s with **at least five** direct reports.
+- The outer query then joins on the `Employee` table to fetch the corresponding manager names.
+
+---
+
+### **Pandas Solution**
+```python
+import pandas as pd
+
+def managers_with_five_reports(employee: pd.DataFrame) -> None:
+    # Count direct reports for each manager
+    report_counts = employee.groupby('managerId').size().reset_index(name='report_count')
+    
+    # Identify managerIds with at least five direct reports
+    valid_managers = report_counts[report_counts['report_count'] >= 5]['managerId']
+    
+    # Filter the Employee table to get manager names
+    # Note: Since managerId can be null, we ignore them during merge.
+    result = employee[employee['id'].isin(valid_managers)][['name']]
+    
+    # Modify the original DataFrame in place if required.
+    employee = result
+    print(result)
+```
+**Explanation:**
+- Group the table by `managerId` and count the number of direct reports.
+- Filter out the managers having at least 5 direct reports.
+- Finally, retrieve the names of these managers.
+
+---
+
+## **File Structure**
+```
+LeetCode570/
+├── problem_statement.md       # Contains the problem description and constraints.
+├── sql_subquery_solution.sql   # SQL solution using subquery.
+├── sql_join_solution.sql       # SQL solution using JOIN.
+├── pandas_solution.py          # Pandas solution for Python users.
+├── README.md                  # Overview of the problem and available solutions.
+```
+
+---
+
+## **Useful Links**
+- [LeetCode Problem 570](https://leetcode.com/problems/managers-with-at-least-5-direct-reports/)
+- [SQL DELETE and JOIN Syntax](https://www.w3schools.com/sql/sql_join.asp)
+- [Pandas Documentation](https://pandas.pydata.org/docs/)
+