LeetCode #184 — MEDIUM

Department Highest Salary

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

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.
Each row of this table indicates the ID, name, and salary of an employee. It also contains the ID of their department.

Table: Department

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
+-------------+---------+
id is the primary key (column with unique values) for this table. It is guaranteed that department name is not NULL.
Each row of this table indicates the ID of a department and its name.

Write a solution to find employees who have the highest salary in each of the departments.

Return the result table in any order.

The result format is in the following example.

Example 1:

Input: 
Employee table:
+----+-------+--------+--------------+
| id | name  | salary | departmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Jim   | 90000  | 1            |
| 3  | Henry | 80000  | 2            |
| 4  | Sam   | 60000  | 2            |
| 5  | Max   | 90000  | 1            |
+----+-------+--------+--------------+
Department table:
+----+-------+
| id | name  |
+----+-------+
| 1  | IT    |
| 2  | Sales |
+----+-------+
Output: 
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Jim      | 90000  |
| Sales      | Henry    | 80000  |
| IT         | Max      | 90000  |
+------------+----------+--------+
Explanation: Max and Jim both have the highest salary in the IT department and Henry has the highest salary in the Sales department.

Step 01

Brute Force Baseline

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

{"headers": {"Employee": ["id", "name", "salary", "departmentId"], "Department": ["id", "name"]}, "rows": {"Employee": [[1, "Joe", 70000, 1], [2, "Jim", 90000, 1], [3, "Henry", 80000, 2], [4, "Sam", 60000, 2], [5, "Max", 90000, 1]], "Department": [[1, "IT"], [2, "Sales"]]}}

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #184: Department Highest Salary
// Auto-generated Java example from py.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (py):
// # Accepted solution for LeetCode #184: Department Highest Salary
// 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

// Accepted solution for LeetCode #184: Department Highest Salary
// Auto-generated Go example from py.
func exampleSolution() {
}
// Reference (py):
// # Accepted solution for LeetCode #184: Department Highest Salary
// 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

# Accepted solution for LeetCode #184: Department Highest Salary
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

// Accepted solution for LeetCode #184: Department Highest Salary
// Rust example auto-generated from py reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (py):
// # Accepted solution for LeetCode #184: Department Highest Salary
// 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

// Accepted solution for LeetCode #184: Department Highest Salary
// Auto-generated TypeScript example from py.
function exampleSolution(): void {
}
// Reference (py):
// # Accepted solution for LeetCode #184: Department Highest Salary
// 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

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.