LeetCode 181: Employees Earning More Than Their Managers ·

Table of Contents

Meta Data
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Difficulty: easy First Attempt: 2021-10-09

Submissions: Runtime Error → Wrong Answer ×2 → Accepted → Wrong Answer → Accepted

Approach
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One table plays both employee and manager, so self-join it (SQL Template 3): e.managerId = m.id pairs each employee with the manager, and e.salary > m.salary filters. Name the output column Employee. People without a manager have a NULL managerId, never join, and are excluded naturally.

My Attempts
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This took 6 submissions:

Attempt 1 selected Name without an alias, ambiguous because both aliases have Name (RE).
With A.Name, the output column was not named Employee; one version switched to SELECT * to inspect the data. a.Name AS 'Employee' passed.

Solution
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SELECT e.name AS Employee
FROM Employee e
JOIN Employee m ON e.managerId = m.id
WHERE e.salary > m.salary;

Complexity
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A self join, near O(n) with an index.

Mistakes to Avoid
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Qualify columns with an alias; a bare Name is ambiguous.
The output column must be named Employee.
Alias case does not matter in MySQL by default, but keep it consistent.

Interview Explanation
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The table plays both roles, so I self-join it, matching each employee’s managerId to the manager’s id, keep employees earning more, and name the column Employee.