Solution Explanation: Managers with at Least 5 Direct Reports

This problem requires finding managers in the Employee table who have at least five direct reports. The solution involves two main steps: counting direct reports for each manager and then filtering to find those with five or more.

Approach:

The most efficient approach uses a combination of GROUP BY, HAVING, and a JOIN operation (or equivalent in other languages like Pandas).

1. Counting Direct Reports: We group the Employee table by managerId. The COUNT(*) aggregate function counts the number of employees for each manager. The HAVING clause filters these groups, keeping only those where the count (COUNT(*)) is greater than or equal to 5. This gives us a table showing managers and their report counts, filtering out those with fewer than five reports.

2. Joining with Employee Table: We then join this intermediate result with the original Employee table using the managerId (from the intermediate table) and the id (from the Employee table). This join links the manager IDs to their corresponding names. We select only the name column from the Employee table to obtain the names of managers who meet the criteria.

Time Complexity Analysis:

• Grouping and Counting: The GROUP BY and COUNT(*) operations have a time complexity of O(N), where N is the number of rows in the Employee table. This is because each row needs to be processed once to assign it to a group and increment the count for that group.

• Filtering (HAVING): The HAVING clause also has a time complexity of O(N) because it needs to iterate through the grouped results (which in the worst case is O(N) groups).

• Joining: The JOIN operation generally has a time complexity of O(Mlog(M) + Nlog(N)) in the best case and O(M*N) in the worst case for a nested loop join, where M is the number of rows in the intermediate result (managers with report counts) and N is the number of rows in the Employee table. However, optimized database systems often use more efficient join algorithms that can significantly improve performance. In practice, with appropriate indexing, the join's complexity might be closer to O(M + N).

Therefore, the overall time complexity of the solution is dominated by the grouping, counting, filtering, and joining steps. The worst-case scenario would result in O(N^2) complexity if no indexes or optimization strategies are employed, However with appropriate indexing, it can reduce the complexity to O(N).

Space Complexity:

The space complexity is determined by the size of the intermediate result (managers with report counts) which is O(M), where M is the number of unique manager IDs. In the worst case, this could be O(N) if every employee is a unique manager. However, in most practical scenarios, the number of managers is significantly smaller than the total number of employees, making the space complexity relatively low.

Code Examples:

MySQL:

SELECT name
FROM Employee
JOIN (
    SELECT managerId, COUNT(*) AS num_reports
    FROM Employee
    GROUP BY managerId
    HAVING num_reports >= 5
) AS ManagerCounts ON Employee.id = ManagerCounts.managerId;

Python (Pandas):

import pandas as pd
 
def find_managers_with_at_least_five_reports(employees: pd.DataFrame) -> pd.DataFrame:
    # Group by managerId and count reports
    manager_reports = employees.groupby('managerId').size().reset_index(name='num_reports')
 
    # Filter for managers with >= 5 reports
    managers_with_five_plus_reports = manager_reports[manager_reports['num_reports'] >= 5]['managerId']
 
    # Merge to get manager names
    result = employees[employees['id'].isin(managers_with_five_plus_reports)][['name']].drop_duplicates()
 
    return result
 

Both examples achieve the same result. The Python example uses Pandas for efficient data manipulation, offering a similar approach to the SQL solution but in a different programming paradigm. The Pandas code is designed for working with dataframes, providing a concise and efficient way to solve the problem in a Python environment.