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Medium

Databases

Employees Earning More Than Their Manager

Problem Context

You are working with a company's HR database. The Employee table stores information about all employees, including their salary and the ID of their manager. Your task is to identify which employees earn a higher salary than their direct manager.

Table Schema

Employee
+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
| salary      | int     |
| managerId   | int     |
+-------------+---------+

id is the primary key.
managerId is a foreign key referencing id. If an employee has no manager, managerId is NULL.

Task

Write a SQL query that returns the names of all employees who earn strictly more than their direct manager. The result table can be returned in any order.

Input / Output Format

Input: The Employee table (provided as a CSV-like representation for testing).

Output: A single-column result named Employee containing the names of qualifying employees.

Example 1

Input Table:

id | name  | salary | managerId
1  | Joe   | 70000  | 3
2  | Henry | 80000  | 4
3  | Sam   | 60000  | NULL
4  | Max   | 90000  | NULL

Output:

Employee
--------
Joe

Explanation: Joe (salary: 70,000) reports to Sam (salary: 60,000). Since 70,000 > 60,000, Joe qualifies. Henry (salary: 80,000) reports to Max (salary: 90,000) — Henry does not qualify.

Example 2

Input Table:

id | name   | salary | managerId
1  | Alice   | 120000 | NULL
2  | Bob     | 130000 | 1
3  | Charlie | 115000 | 1

Output:

Employee
--------
Bob

Explanation: Both Bob and Charlie report to Alice (120,000). Bob (130,000) earns more, Charlie (115,000) does not.

Approach Hints

Self-Join: Join the Employee table with itself — one alias for the employee, another for the manager — linking on employee.managerId = manager.id.
Filter Condition: Apply a WHERE clause comparing employee.salary > manager.salary.
NULL Handling: Employees with NULL managerId are top-level executives with no manager; they should be automatically excluded by an INNER JOIN since there is no matching manager row.

Constraints

1 <= number of rows in Employee <= 10^4
1 <= id <= 10^4
1 <= salary <= 10^6
managerId is either NULL or a valid id in the Employee table
All employee names are non-empty strings of length <= 255
No two employees share the same id

Examples

Input: Employee table: id=1, name='Joe', salary=70000, managerId=3 id=2, name='Henry', salary=80000, managerId=4 id=3, name='Sam', salary=60000, managerId=NULL id=4, name='Max', salary=90000, managerId=NULL

Output: Employee Joe

Joe earns 70,000 and his manager Sam earns 60,000. Since 70,000 > 60,000, Joe is included. Henry earns 80,000 but his manager Max earns 90,000, so Henry is excluded. Sam and Max have no managers and are excluded.

Input: Employee table: id=1, name='Alice', salary=120000, managerId=NULL id=2, name='Bob', salary=130000, managerId=1 id=3, name='Charlie', salary=115000, managerId=1

Output: Employee Bob

Bob earns 130,000 vs Alice's 120,000 — qualifies. Charlie earns 115,000 vs Alice's 120,000 — does not qualify.

Input: Employee table: id=1, name='CEO', salary=200000, managerId=NULL id=2, name='VP', salary=150000, managerId=1 id=3, name='Manager', salary=100000, managerId=2 id=4, name='Engineer', salary=160000, managerId=3

Output: Employee Engineer

CEO has no manager. VP (150k) < CEO (200k). Manager (100k) < VP (150k). Engineer (160k) > Manager (100k), so Engineer qualifies.

Sample Test Cases

Case 1 Input: id=1, name='Joe', salary=70000, managerId=3 id=2, name='Henry', salary=80000, managerId=4 id=3, name='Sam', salary=60000, managerId=NULL id=4, name='Max', salary=90000, managerId=NULL

Expected: Employee Joe

Case 2 Input: id=1, name='Alice', salary=120000, managerId=NULL id=2, name='Bob', salary=130000, managerId=1 id=3, name='Charlie', salary=115000, managerId=1

Expected: Employee Bob

Case 3 Input: id=1, name='CEO', salary=200000, managerId=NULL id=2, name='VP', salary=150000, managerId=1 id=3, name='Manager', salary=100000, managerId=2 id=4, name='Engineer', salary=160000, managerId=3

Expected: Employee Engineer

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Case 1 Input: id=1, name='Joe', salary=70000, managerId=3 id=2, name='Henry', salary=80000, managerId=4 id=3, name='Sam', salary=60000, managerId=NULL id=4, name='Max', salary=90000, managerId=NULL

Expected: Employee Joe

Case 2 Input: id=1, name='Alice', salary=120000, managerId=NULL id=2, name='Bob', salary=130000, managerId=1 id=3, name='Charlie', salary=115000, managerId=1

Expected: Employee Bob

Expected: Employee Engineer

Table Schema

Employee +-------------+---------+ | Column Name | Type | +-------------+---------+ | id | int | | name | varchar | | salary | int | | managerId | int | +-------------+---------+

id is the primary key.

managerId is a foreign key referencing id. If an employee has no manager, managerId is NULL.

Example 1

Input Table:

id | name | salary | managerId 1 | Joe | 70000 | 3 2 | Henry | 80000 | 4 3 | Sam | 60000 | NULL 4 | Max | 90000 | NULL

Output:

Employee -------- Joe

Explanation: Joe (salary: 70,000) reports to Sam (salary: 60,000). Since 70,000 > 60,000, Joe qualifies. Henry (salary: 80,000) reports to Max (salary: 90,000) — Henry does not qualify.

Approach Hints

Self-Join: Join the Employee table with itself — one alias for the employee, another for the manager — linking on employee.managerId = manager.id.

Filter Condition: Apply a WHERE clause comparing employee.salary > manager.salary.

NULL Handling: Employees with NULL managerId are top-level executives with no manager; they should be automatically excluded by an INNER JOIN since there is no matching manager row.

1 <= number of rows in Employee <= 10^4 1 <= id <= 10^4 1 <= salary <= 10^6 managerId is either NULL or a valid id in the Employee table All employee names are non-empty strings of length <= 255 No two employees share the same id