Solution Explanation: Counting Students per Department

This problem requires retrieving the number of students in each department, including departments with no students. The solution uses SQL to efficiently handle this.

Approach

The core idea is to perform a LEFT JOIN between the Department and Student tables. A LEFT JOIN ensures that all rows from the left table (here, Department) are included in the result, even if there's no matching row in the right table (Student).

1. LEFT JOIN: We join Department and Student tables using dept_id as the common key. This links departments to their students. The USING(dept_id) clause is a shorthand for specifying the join condition.

2. COUNT(student_id): We use COUNT(student_id) to count the number of students in each department. Crucially, if a department has no students, COUNT(student_id) will return 0 for that department.

3. GROUP BY dept_id: This groups the results by department ID, so we get a single count for each department.

4. ORDER BY 2 DESC, 1: This sorts the result set. It first sorts by the second column (student_number) in descending order (departments with more students appear first). If there's a tie in student count, it then sorts by the first column (dept_name) in ascending order (alphabetical order).

SQL Code (MySQL)

SELECT dept_name, COUNT(student_id) AS student_number
FROM
    Department
    LEFT JOIN Student USING (dept_id)
GROUP BY dept_id
ORDER BY 2 DESC, 1;

Time Complexity Analysis

• LEFT JOIN: The time complexity of a LEFT JOIN operation generally depends on the size of the tables and the indexing. In the best case (with appropriate indexes on dept_id), it can be close to O(m + n), where 'm' is the number of rows in Department and 'n' is the number of rows in Student. In the worst case (no suitable indexes), it could approach O(m*n).

• GROUP BY: The time complexity of the GROUP BY operation is typically O(N log N) or O(N), depending on the specific database implementation and the presence of indexes. 'N' here refers to the total number of rows after the join.

• ORDER BY: The ORDER BY operation generally has a time complexity of O(N log N) for sorting N rows.

Therefore, the overall time complexity of the query is dominated by the JOIN, GROUP BY, and ORDER BY operations. The worst-case complexity is O(m*n) if there are no appropriate indexes, but with proper indexing, it's much closer to O(N log N), where N is at most m+n.

Space Complexity Analysis

The space complexity is primarily determined by the size of the intermediate result set after the LEFT JOIN and GROUP BY operations. In the worst case, this could be proportional to the total number of rows in the Department and Student tables, which we can represent as O(m + n). The final result set will be smaller (at most the number of departments).