Merge pull request #73 from iamAntimPal/Branch-1

Branch 1

Author: iamAntimPal

LeetCode SQL 50 Solution/185. Department Top Three Salaries/185. Department Top Three Salaries.md

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+### **Pandas Solution**
+
+import pandas as pd
+
+def department_top_three_salaries(employee: pd.DataFrame, department: pd.DataFrame) -> pd.DataFrame:
+    # Merge employee and department tables
+    employee = employee.merge(department, left_on='departmentId', right_on='id', suffixes=('', '_dept'))
+    
+    # Rank employees' salaries within each department
+    employee['rank'] = employee.groupby('departmentId')['salary'].rank(method='dense', ascending=False)
+    
+    # Filter top 3 salaries in each department
+    result = employee[employee['rank'] <= 3][['name_dept', 'name', 'salary']]
+    
+    # Rename columns to match the expected output
+    result.columns = ['Department', 'Employee', 'Salary']
+    
+    return result

LeetCode SQL 50 Solution/185. Department Top Three Salaries/185. Department Top Three Salaries.py

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+# **185. Department Top Three Salaries**
+
+## **Problem Statement**
+You are given two tables: `Employee` and `Department`.
+
+### **Employee Table**
+```rb
++--------------+---------+
+| Column Name  | Type    |
++--------------+---------+
+| id           | int     |
+| name         | varchar |
+| salary       | int     |
+| departmentId | int     |
++--------------+---------+
+```
+- `id` is the primary key.
+- `departmentId` is a foreign key referencing `id` in the `Department` table.
+- Each row represents an employee with their `id`, `name`, `salary`, and `departmentId`.
+
+### **Department Table**
+```rb
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| id          | int     |
+| name        | varchar |
++-------------+---------+
+```
+- `id` is the primary key.
+- Each row represents a department with its `id` and `name`.
+
+### **Task:**
+Find employees who have a salary in the **top three unique salaries** in their respective departments.
+
+## **Example 1:**
+### **Input:**
+#### **Employee Table**
+```
++----+-------+--------+--------------+
+| id | name  | salary | departmentId |
++----+-------+--------+--------------+
+| 1  | Joe   | 85000  | 1            |
+| 2  | Henry | 80000  | 2            |
+| 3  | Sam   | 60000  | 2            |
+| 4  | Max   | 90000  | 1            |
+| 5  | Janet | 69000  | 1            |
+| 6  | Randy | 85000  | 1            |
+| 7  | Will  | 70000  | 1            |
++----+-------+--------+--------------+
+```
+#### **Department Table**
+```rb
++----+-------+
+| id | name  |
++----+-------+
+| 1  | IT    |
+| 2  | Sales |
++----+-------+
+```
+### **Output:**
+```rb
++------------+----------+--------+
+| Department | Employee | Salary |
++------------+----------+--------+
+| IT         | Max      | 90000  |
+| IT         | Joe      | 85000  |
+| IT         | Randy    | 85000  |
+| IT         | Will     | 70000  |
+| Sales      | Henry    | 80000  |
+| Sales      | Sam      | 60000  |
++------------+----------+--------+
+```
+
+---
+
+## **Solution Approaches**
+
+### **SQL Solution (Using Self Join)**
+```sql
+SELECT d.Name as Department,
+       e.Name as Employee,
+       e.Salary as Salary
+FROM Department d, Employee e
+WHERE (
+    SELECT COUNT(DISTINCT Salary)
+    FROM Employee
+    WHERE Salary > e.Salary AND DepartmentId = d.Id
+) < 3 AND e.DepartmentId = d.Id
+ORDER BY d.Id, e.Salary DESC;
+```
+**Explanation:**
+- For each employee, we count how many distinct salaries are greater than theirs.
+- If fewer than 3 salaries are greater, the employee is in the **top three**.
+- We filter results by department and order by salary in descending order.
+
+---
+
+### **SQL Solution (Using Window Functions)**
+```sql
+WITH RankedSalaries AS (
+    SELECT e.name AS Employee, 
+           e.salary AS Salary, 
+           d.name AS Department,
+           DENSE_RANK() OVER (PARTITION BY e.departmentId ORDER BY e.salary DESC) AS rnk
+    FROM Employee e
+    JOIN Department d ON e.departmentId = d.id
+)
+SELECT Department, Employee, Salary
+FROM RankedSalaries
+WHERE rnk <= 3;
+```
+**Explanation:**
+- We use `DENSE_RANK()` to assign a rank to salaries within each department.
+- `PARTITION BY departmentId` ensures ranking is specific to each department.
+- Employees with `rnk <= 3` are returned.
+
+---
+
+### **Pandas Solution**
+```python
+import pandas as pd
+
+def department_top_three_salaries(employee: pd.DataFrame, department: pd.DataFrame) -> pd.DataFrame:
+    # Merge employee and department tables
+    employee = employee.merge(department, left_on='departmentId', right_on='id', suffixes=('', '_dept'))
+    
+    # Rank employees' salaries within each department
+    employee['rank'] = employee.groupby('departmentId')['salary'].rank(method='dense', ascending=False)
+    
+    # Filter top 3 salaries in each department
+    result = employee[employee['rank'] <= 3][['name_dept', 'name', 'salary']]
+    
+    # Rename columns to match the expected output
+    result.columns = ['Department', 'Employee', 'Salary']
+    
+    return result
+```
+**Explanation:**
+- Merge the `Employee` and `Department` tables.
+- Rank salaries within each department using `.rank()`.
+- Filter the top 3 ranked salaries per department.
+
+---
+
+## **File Structure**
+```
+📂 LeetCode185
+│── 📜 problem_statement.md
+│── 📜 sql_self_join_solution.sql
+│── 📜 sql_window_function_solution.sql
+│── 📜 pandas_solution.py
+│── 📜 README.md
+```
+- `problem_statement.md` → Contains the problem description and constraints.
+- `sql_self_join_solution.sql` → Contains the SQL solution using self-join.
+- `sql_window_function_solution.sql` → Contains the SQL solution using `DENSE_RANK()`.
+- `pandas_solution.py` → Contains the Pandas solution for Python users.
+- `README.md` → Provides an overview of the problem and solutions.
+
+---
+
+## **Useful Links**
+- [LeetCode Problem 185](https://leetcode.com/problems/department-top-three-salaries/)
+- [SQL DENSE_RANK() Function](https://www.w3schools.com/sql/sql_functions.asp)
+- [Pandas Rank Documentation](https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.rank.html)
+
+---
+

LeetCode SQL 50 Solution/185. Department Top Three Salaries/readme.md

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+import pandas as pd
+
+def delete_duplicate_emails(person: pd.DataFrame) -> None:
+    # Keep only the first occurrence of each email (smallest id)
+    person.drop_duplicates(subset=['email'], keep='first', inplace=True)

LeetCode SQL 50 Solution/196. Delete Duplicate Emails/196. Delete Duplicate Emails.py

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-196. Delete Duplicate Emails
-Solved
-Easy
-Topics
-Companies
-SQL Schema
-Pandas Schema
-Table: Person
-
-+-------------+---------+
-| Column Name | Type    |
-+-------------+---------+
-| id          | int     |
-| email       | varchar |
-+-------------+---------+
-id is the primary key (column with unique values) for this table.
-Each row of this table contains an email. The emails will not contain uppercase letters.
- 
-
-Write a solution to delete all duplicate emails, keeping only one unique email with the smallest id.
-
-For SQL users, please note that you are supposed to write a DELETE statement and not a SELECT one.
-
-For Pandas users, please note that you are supposed to modify Person in place.
-
-After running your script, the answer shown is the Person table. The driver will first compile and run your piece of code and then show the Person table. The final order of the Person table does not matter.
-
-The result format is in the following example.
-
- 
-
-Example 1:
-
-Input: 
-Person table:
-+----+------------------+
-| id | email            |
-+----+------------------+
-| 1  | john@example.com |
-| 2  | bob@example.com  |
-| 3  | john@example.com |
-+----+------------------+
-Output: 
-+----+------------------+
-| id | email            |
-+----+------------------+
-| 1  | john@example.com |
-| 2  | bob@example.com  |
-+----+------------------+
-Explanation: john@example.com is repeated two times. We keep the row with the smallest Id = 1.
-
-
-
+-- 196. Delete Duplicate Emails
 # Write your MySQL query statement below
 # Write your MySQL query statement below
 DELETE p2 FROM Person p1