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Medium

Databases

Employees Earning More Than Their Manager

Problem Context

You are working with a company's HR database and need to identify employees who are earning a higher salary than the manager they directly report to. This is a classic SQL self-join problem that tests your ability to query hierarchical data within a single table.

Table Schema

You are given a single table called Employee with the following structure:

Column Name	Type	Description
id	int	Primary key, unique employee identifier
name	varchar	Employee's full name
salary	int	Employee's annual salary
managerId	int	Foreign key referencing the `id` of their manager (NULL if no manager)

Task

Write a SQL query to find the names of all employees who earn strictly more than their direct manager. Return the result in any order.

Input/Output Format

Input: The Employee table as described above.

Output: A result set with a single column Employee containing the names of qualifying employees.

Examples

Example 1

Employee 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: 70000) reports to Sam (salary: 60000). Since 70000 > 60000, Joe qualifies. Henry (salary: 80000) reports to Max (salary: 90000). Since 80000 < 90000, Henry does NOT qualify.

Approach Hints

Self-Join Approach: Join the Employee table with itself — once as the employee (e1) and once as the manager (e2) — matching on e1.managerId = e2.id. Then filter rows where e1.salary > e2.salary.
Subquery Approach: For each employee who has a manager, use a correlated subquery to fetch their manager's salary and compare.
Key Considerations:

Constraints

- The `Employee` table may have 0 to 10,000 rows.
- `id` is unique and non-null.
- `salary` values are in the range `[1000, 1,000,000]`.
- `managerId` is either NULL or a valid `id` within the same table.
- No circular management chains (an employee cannot be their own manager, directly or indirectly).
- Employee names are non-empty strings of up to 50 characters.

Examples

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

Output: | Employee | |----------| | Joe |

Joe earns 70000 and his manager Sam earns 60000. 70000 > 60000 so Joe qualifies. Henry earns 80000 and his manager Max earns 90000. 80000 < 90000 so Henry does not qualify. Sam and Max have no managers and are excluded.

Input: Employee table: | id | name | salary | managerId | |----|---------|--------|-----------| | 1 | Alice | 120000 | 2 | | 2 | Bob | 100000 | NULL | | 3 | Charlie | 95000 | 2 | | 4 | Diana | 110000 | 1 |

Output: | Employee | |----------| | Alice | | Diana |

Alice (120000) reports to Bob (100000) — 120000 > 100000, qualifies. Charlie (95000) reports to Bob (100000) — 95000 < 100000, does not qualify. Diana (110000) reports to Alice (120000) — 110000 < 120000, does not qualify. Wait — Diana (110000) reports to Alice (120000)... actually Diana does NOT qualify here. Let me recheck: Alice qualifies (120k > 100k Bob). Diana (110k) < Alice (120k), so Diana does NOT qualify. Only Alice qualifies.

Input: Employee table: | id | name | salary | managerId | |----|-------|--------|-----------| | 1 | Eve | 50000 | NULL |

Output: | Employee | |----------|

There is only one employee with no manager. No one can qualify since there are no manager-employee pairs to compare. The result is an empty table.

Sample Test Cases

Case 1 Input: Employee table: | id | name | salary | managerId | |----|-------|--------|-----------| | 1 | Joe | 70000 | 3 | | 2 | Henry | 80000 | 4 | | 3 | Sam | 60000 | NULL | | 4 | Max | 90000 | NULL |

Expected: Joe

Case 2 Input: Employee table: | id | name | salary | managerId | |----|---------|--------|-----------| | 1 | Alice | 120000 | 2 | | 2 | Bob | 100000 | NULL | | 3 | Charlie | 95000 | 2 |

Expected: Alice

Case 3 Input: Employee table: | id | name | salary | managerId | |----|-------|--------|-----------| | 1 | Eve | 50000 | NULL |

Expected:

Loading editor...

Expected: Joe

Case 2 Input: Employee table: | id | name | salary | managerId | |----|---------|--------|-----------| | 1 | Alice | 120000 | 2 | | 2 | Bob | 100000 | NULL | | 3 | Charlie | 95000 | 2 |

Expected: Alice

Case 3 Input: Employee table: | id | name | salary | managerId | |----|-------|--------|-----------| | 1 | Eve | 50000 | NULL |

Expected:

Column Name

Type

Description

int

Primary key, unique employee identifier

name

varchar

Employee's full name

salary

int

Employee's annual salary

managerId

int

Foreign key referencing the id of their manager (NULL if no manager)

Examples

Example 1

Employee Table:

name

salary

managerId

Joe

70000

Henry

80000

Sam

60000

NULL

Max

90000

NULL

Output:

Employee

Joe

Approach Hints

Self-Join Approach: Join the Employee table with itself — once as the employee (e1) and once as the manager (e2) — matching on e1.managerId = e2.id. Then filter rows where e1.salary > e2.salary.

Subquery Approach: For each employee who has a manager, use a correlated subquery to fetch their manager's salary and compare.

Key Considerations:

- The `Employee` table may have 0 to 10,000 rows. - `id` is unique and non-null. - `salary` values are in the range `[1000, 1,000,000]`. - `managerId` is either NULL or a valid `id` within the same table. - No circular management chains (an employee cannot be their own manager, directly or indirectly). - Employee names are non-empty strings of up to 50 characters.