Solution Explanation and Code

This problem requires finding the projects with the maximum number of employees. We can achieve this using SQL queries. Two approaches are presented below: one using a HAVING clause with a subquery and another using window functions with RANK().

Solution 1: Using HAVING Clause and Subquery

This approach first groups the Project table by project_id to count the number of employees for each project. Then, the HAVING clause filters these groups, selecting only those projects where the employee count (COUNT(1)) is greater than or equal to the maximum employee count across all projects. The subquery (SELECT COUNT(1) FROM Project GROUP BY project_id) finds the counts for each project. >= ALL ensures that only projects with the highest employee count are returned.

MySQL Code:

SELECT project_id
FROM Project
GROUP BY 1
HAVING
    COUNT(1) >= all(
        SELECT COUNT(1)
        FROM Project
        GROUP BY project_id
    );

Time Complexity: The time complexity of this query is dominated by the grouping and counting operations. In the worst case, where the number of projects is n, the time complexity is O(n log n) due to sorting implicitly involved in finding the maximum. However, database engines are highly optimized, and the actual performance may vary depending on indexes and database specifics.

Solution 2: Using Window Function RANK()

This approach uses a window function to assign a rank to each project based on the number of employees. The RANK() function assigns the same rank to projects with the same number of employees. A Common Table Expression (CTE) called T is used to perform the ranking. Then, a final SELECT statement filters the results from T to select only projects with rank 1 (projects with the maximum number of employees).

MySQL Code:

WITH
    T AS (
        SELECT
            project_id,
            RANK() OVER (ORDER BY COUNT(employee_id) DESC) AS rk
        FROM Project
        GROUP BY 1
    )
SELECT project_id
FROM T
WHERE rk = 1;

Time Complexity: The COUNT(employee_id) and GROUP BY operations take O(n) time, where n is the number of rows in the Project table. The RANK() window function typically has a time complexity of O(n log n) because of sorting (though database implementations may optimize this). The final SELECT operation is O(n) in the worst case. Therefore, the overall time complexity is dominated by the ranking and is O(n log n). Again, database optimizations can significantly affect the actual performance.

Comparison of Solutions:

Both solutions achieve the same outcome. The second solution using window functions is often considered more readable and potentially more efficient in some database systems, as window functions are designed for this type of ranking and aggregation. However, the performance difference might be negligible for smaller datasets. The choice depends on personal preference and database system capabilities.