Watch the video solution for Manager of the Largest Department, a medium level problem involving Database. This walkthrough by Everyday Data Science has 501 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
Table: Employees
+-------------+---------+ | Column Name | Type | +-------------+---------+ | emp_id | int | | emp_name | varchar | | dep_id | int | | position | varchar | +-------------+---------+ emp_id is column of unique values for this table. This table contains emp_id, emp_name, dep_id, and position.
Write a solution to find the name of the manager from the largest department. There may be multiple largest departments when the number of employees in those departments is the same.
Return the result table sorted by dep_id in ascending order.
The result format is in the following example.
Example 1:
Input: Employees table: +--------+----------+--------+---------------+ | emp_id | emp_name | dep_id | position | +--------+----------+--------+---------------+ | 156 | Michael | 107 | Manager | | 112 | Lucas | 107 | Consultant | | 8 | Isabella | 101 | Manager | | 160 | Joseph | 100 | Manager | | 80 | Aiden | 100 | Engineer | | 190 | Skylar | 100 | Freelancer | | 196 | Stella | 101 | Coordinator | | 167 | Audrey | 100 | Consultant | | 97 | Nathan | 101 | Supervisor | | 128 | Ian | 101 | Administrator | | 81 | Ethan | 107 | Administrator | +--------+----------+--------+---------------+ Output +--------------+--------+ | manager_name | dep_id | +--------------+--------+ | Joseph | 100 | | Isabella | 101 | +--------------+--------+ Explanation - Departments with IDs 100 and 101 each has a total of 4 employees, while department 107 has 3 employees. Since both departments 100 and 101 have an equal number of employees, their respective managers will be included. Output table is ordered by dep_id in ascending order.
Problem Overview: The query asks for the manager responsible for the department that has the highest number of employees. You first need to count how many employees belong to each department, then identify the department with the largest count, and finally return the manager of that department.
Approach 1: Grouping + Equi-Join + Subquery (O(n) time, O(1) extra space)
This approach relies on SQL aggregation to measure department size. Start by grouping employees by department_id and computing COUNT(*) to determine how many employees belong to each department. A subquery then finds the maximum department size using MAX() over those grouped counts. The outer query joins the department table with the grouped results using an equi-join and filters for the department whose count equals the maximum.
The key insight is separating the problem into two stages: compute department sizes, then select the largest one. SQL handles this efficiently with GROUP BY and a nested subquery. The equi-join ensures you retrieve the manager assigned to that department. This pattern is common in SQL interview questions where you must combine aggregation with relational joins.
If multiple departments tie for the largest size, the query naturally returns all matching managers because the filter compares counts against the maximum value. This makes the solution robust without additional conditional logic. The database engine performs a scan of the employee table, builds grouped counts, and evaluates the maximum in the subquery.
Conceptually, the solution combines three core database operations: aggregation (GROUP BY), relational matching via joins, and filtering with a scalar subquery. These patterns appear frequently in SQL interview problems involving "largest", "highest", or "top" entity queries.
Recommended for interviews: The grouping + subquery solution is the expected approach. It clearly demonstrates understanding of aggregation and relational joins. Variants using window functions or ordering with LIMIT may also work, but the explicit GROUP BY + MAX() subquery shows the strongest command of core SQL fundamentals.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Grouping + Subquery + Join | O(n) | O(1) | Standard SQL solution for finding entities with maximum aggregated values |
| GROUP BY + ORDER BY + LIMIT | O(n log n) | O(1) | Useful when only the top department is required and sorting is acceptable |
| Window Function (RANK / DENSE_RANK) | O(n) | O(n) | Modern SQL engines where analytic functions are preferred |
