adding pandas solution on 184 Leetcode question

Author: iamAntimPal

solution/0100-0199/0184.Department Highest Salary/README.md

@@ -92,6 +92,28 @@ Department 表:
 
 <!-- solution:start -->
 
+### Python3
+```python
+import pandas as pd
+
+def department_highest_salary(employee: pd.DataFrame, department: pd.DataFrame) -> pd.DataFrame:
+    # Merge the two tables on departmentId and department id
+    merged = employee.merge(department, left_on='departmentId', right_on='id')
+    
+    # Find the maximum salary for each department
+    max_salaries = merged.groupby('departmentId')['salary'].transform('max')
+    
+    # Filter employees who have the highest salary in their department
+    top_earners = merged[merged['salary'] == max_salaries]
+    
+    # Select required columns and rename them
+    result = top_earners[['name_y', 'name_x', 'salary']].copy()
+    result.columns = ['Department', 'Employee', 'Salary']
+    
+    return result
+```
+
+
 ### 方法一：等值连接 + 子查询
 
 我们可以使用等值连接，将 `Employee` 表和 `Department` 表连接起来，连接条件为 `Employee.departmentId = Department.id`，然后使用子查询来找到每个部门的最高工资，最后使用 `WHERE` 子句来筛选出每个部门中薪资最高的员工。

solution/0100-0199/0184.Department Highest Salary/README_EN.md

@@ -94,6 +94,28 @@ Department table:
 
 <!-- solution:start -->
 
+### Python3
+```python
+import pandas as pd
+
+def department_highest_salary(employee: pd.DataFrame, department: pd.DataFrame) -> pd.DataFrame:
+    # Merge the two tables on departmentId and department id
+    merged = employee.merge(department, left_on='departmentId', right_on='id')
+    
+    # Find the maximum salary for each department
+    max_salaries = merged.groupby('departmentId')['salary'].transform('max')
+    
+    # Filter employees who have the highest salary in their department
+    top_earners = merged[merged['salary'] == max_salaries]
+    
+    # Select required columns and rename them
+    result = top_earners[['name_y', 'name_x', 'salary']].copy()
+    result.columns = ['Department', 'Employee', 'Salary']
+    
+    return result
+```
+
+
 ### Solution 1: Equi-Join + Subquery
 
 We can use an equi-join to join the `Employee` table and the `Department` table based on `Employee.departmentId = Department.id`, and then use a subquery to find the highest salary for each department. Finally, we can use a `WHERE` clause to filter out the employees with the highest salary in each department.
@@ -151,6 +173,8 @@ FROM T
 WHERE rk = 1;
 ```
 
+
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0100-0199/0184.Department Highest Salary/solution.py

@@ -0,0 +1,17 @@
+import pandas as pd
+
+def department_highest_salary(employee: pd.DataFrame, department: pd.DataFrame) -> pd.DataFrame:
+    # Merge the two tables on departmentId and department id
+    merged = employee.merge(department, left_on='departmentId', right_on='id')
+    
+    # Find the maximum salary for each department
+    max_salaries = merged.groupby('departmentId')['salary'].transform('max')
+    
+    # Filter employees who have the highest salary in their department
+    top_earners = merged[merged['salary'] == max_salaries]
+    
+    # Select required columns and rename them
+    result = top_earners[['name_y', 'name_x', 'salary']].copy()
+    result.columns = ['Department', 'Employee', 'Salary']
+    
+    return result