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This approach uses SQL's GROUP BY and HAVING clause to count the number of employees managed by each manager. We group the employees by their managerId and count the occurrences to determine the number of direct reports. We then filter the groups having at least five direct reports and join back with the Employee table to get the names of these managers.
The time complexity of this query depends on the indexing and size of the original table. In general, it's O(n) for counting and filtering operations, where n is the number of rows in the Employee table.
1SELECT e.name FROM Employee e WHERE e.id IN (SELECT managerId FROM Employee WHERE managerId IS NOT NULL GROUP BY managerId HAVING COUNT(id) >= 5);First, we select the managers who have at least five direct reports by counting the number of employees corresponding to each managerId in the subquery. We then use this list to filter the original table and retrieve the names of these managers from the Employee table.
This approach utilizes a join between the table itself, followed by a GROUP BY and filter with a HAVING clause. We join the Employee table with itself on the condition of matching the managerId to id. This join helps to create pairs of managers and their direct reports. By grouping based on the manager's id and applying a HAVING clause, we fetch managers with five or more reports.
The time complexity for this query is similar to the previous approach, influenced by both the number of rows to join and then group. Generally, it would be about O(n) constrained within the database engine optimizations.
1SELECT e1.name FROM Employee e1 JOIN Employee e2 ON e1.id = e2.managerId GROUP BYThis approach involves using a SQL query to group employees by their managerId and then counting how many employees report to each manager. From this, we can filter out managers who have less than five direct reports.
Time Complexity: O(n), where n is the number of employees as we scan the list and group it.
Space Complexity: O(m), where m is the number of unique managerId entries that meet the criteria.
1SELECT e.name
2FROM Employee e
3JOIN (SELECT managerId
4 This solution uses a self join within the table and counts the direct reports for each manager by referring to the managerId.
Time Complexity: O(n), due to the need to scan and join the relationships within the table.
Space Complexity: O(m), where m is the number of distinct managers in the table.
1SELECT e1.name
2FROM Employee e1
3JOIN Employee e2 ON e1.id = e2.managerId
4In this implementation, we first join the table Employee with itself to create pairs between managers and their direct reports. We then group by the manager's id, counting the number of reports for each manager to filter those with at least five.
We first create a subquery to select managerId from the Employee table, grouping by managerId and including only those managerIds with five or more entries. Then, we join this result with the Employee table to get the names of these managers.
Here, we perform a self join on the Employee table. We join on e1.id and e2.managerId, essentially matching each manager with their subordinates. We then group by the manager's id to count how many subordinates each has, and filter to keep those with at least 5 subordinates.