Leetcode #184: Department Highest Salary
In this guide, we solve Leetcode #184 Department Highest Salary in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Table: Employee +--------------+---------+ | Column Name | Type | +--------------+---------+ | id | int | | name | varchar | | salary | int | | departmentId | int | +--------------+---------+ id is the primary key (column with unique values) for this table. departmentId is a foreign key (reference columns) of the ID from the Department table.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Database
Intuition
The task is relational in nature, which maps cleanly to DataFrame operations in Python.
By treating tables as DataFrames, joins and group-bys become concise and readable.
Approach
Load the inputs as DataFrames and apply the appropriate merge, filter, or group-by.
Select or rename the columns to match the required output.
Steps:
- Load inputs as DataFrames.
- Apply merge/groupby/filter operations.
- Select the output columns.
Example
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| id | int |
| name | varchar |
| salary | int |
| departmentId | int |
+--------------+---------+
id is the primary key (column with unique values) for this table.
departmentId is a foreign key (reference columns) of the ID from the Department table.
Each row of this table indicates the ID, name, and salary of an employee. It also contains the ID of their department.
Python Solution
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
Complexity
The time complexity is O(n log n) (typical). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.