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Employee and manager names

SQL data engineering interview problem. Difficulty: intermediate. Pattern: Self Joins. About 12 minutes. Part of the Pro drill bank.

Using a self-join on employees, return each employee who has a manager. Columns: employee_name, manager_name. Order by employee_name.

Requirements

  • Only employees with a non-null manager_id appear.

Examples

Input: employees employee_id | name | manager_id 1 | Alice Chen | NULL 2 | Bob Kumar | 1 4 | Eve Ng | 2 5 | Omar Ali | 1 Output: employee_name | manager_name Bob Kumar | Alice Chen Eve Ng | Bob Kumar Omar Ali | Alice Chen Why this passes: Alice has no manager_id so the join drops her.

Topics: lakebench, sql, self-join.

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intermediate

Employee and manager names

Interview-style drill: Self-join employees to resolve manager names.

Using a self-join on `employees`, return each employee who has a manager. Columns: `employee_name`, `manager_name`. Order by `employee_name`.