LeetCode #185 — HARD

Department Top Three Salaries

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

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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 column) 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.
Each row of this table indicates the ID of a department and its name.

A company's executives are interested in seeing who earns the most money in each of the company's departments. A high earner in a department is an employee who has a salary in the top three unique salaries for that department.

Write a solution to find the employees who are high earners 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   | 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:
+----+-------+
| id | name  |
+----+-------+
| 1  | IT    |
| 2  | Sales |
+----+-------+
Output: 
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Joe      | 85000  |
| IT         | Randy    | 85000  |
| IT         | Will     | 70000  |
| Sales      | Henry    | 80000  |
| Sales      | Sam      | 60000  |
+------------+----------+--------+
Explanation: 
In the IT department:
- Max earns the highest unique salary
- Both Randy and Joe earn the second-highest unique salary
- Will earns the third-highest unique salary

In the Sales department:
- Henry earns the highest salary
- Sam earns the second-highest salary
- There is no third-highest salary as there are only two employees

Constraints:

  • There are no employees with the exact same name, salary and department.

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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 column) 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. Each row of this table indicates the ID of a department and its name. A company's executives are interested in seeing who earns the most money in each of the company's departments. A high earner in a department is an

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", 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": [[1, "IT"], [2, "Sales"]]}}
Step 02

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

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #185: Department Top Three Salaries
// Auto-generated Java example from py.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (py):
// # Accepted solution for LeetCode #185: Department Top Three Salaries
// import pandas as pd
// 
// 
// def top_three_salaries(
//     employee: pd.DataFrame, department: pd.DataFrame
// ) -> pd.DataFrame:
//     salary_cutoff = (
//         employee.drop_duplicates(["salary", "departmentId"])
//         .groupby("departmentId")["salary"]
//         .nlargest(3)
//         .groupby("departmentId")
//         .min()
//     )
//     employee["Department"] = department.set_index("id")["name"][
//         employee["departmentId"]
//     ].values
//     employee["cutoff"] = salary_cutoff[employee["departmentId"]].values
//     return employee[employee["salary"] >= employee["cutoff"]].rename(
//         columns={"name": "Employee", "salary": "Salary"}
//     )[["Department", "Employee", "Salary"]]
Step 06

Interactive Study Demo

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

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.