How do you solve Students and Examinations on LeetCode?

Students and Examinations (LeetCode #1280) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how joins work in SQL is crucial for efficient data retrieval.

What algorithm or data structure is used to solve Students and Examinations?

Students and Examinations uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Students and Examinations?

Always clarify the requirements and constraints of the problem. Practice writing SQL queries on paper to improve your fluency. Understand the underlying data model and relationships between tables.

What are common mistakes when solving Students and Examinations?

Not using GROUP BY correctly, leading to incorrect counts. Forgetting to handle NULL values when using LEFT JOIN.

Students and Examinations — LeetCode #1280 (Easy)

Tags: Database

Related patterns: Hash Map, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

In the brute force approach, we will iterate through each student and each subject, counting how many times each student attended each exam by checking the Examinations table. This is straightforward but inefficient as it involves nested loops.

The time complexity is O(n²) because we are iterating through each student and each subject, leading to a nested loop structure. The space complexity is O(1) as we are not using additional data structures that grow with input size.

Step 1: Initialize an empty result list to store the counts.
Step 2: For each student in the Students table, iterate through each subject in the Subjects table.
Step 3: For each student-subject pair, count occurrences in the Examinations table and store the result.

1. Students: [(1, Alice), (2, Bob), (13, John), (6, Alex)], Subjects: [(Math), (Science)], Examinations: [(1, Math), (1, Science), (2, Math)] 2. For student 1 (Alice) and subject Math, count = 1 (found in Examinations) 3. For student 1 (Alice) and subject Science, count = 1 (found in Examinations) 4. For student 2 (Bob) and subject Math, count = 1 (found in Examinations) 5. For student 2 (Bob) and subject Science, count = 0 (not found) 6. Continue for other students and subjects.

Optimal Solution approach

Time complexity: O(n). Space complexity: O(n).

The optimal solution leverages SQL's aggregation functions and joins to efficiently count the number of times each student attended each exam without the need for nested loops. This reduces the time complexity significantly.

The time complexity is O(n) because we are performing a single pass through the data with joins and aggregations. The space complexity is O(n) due to the storage of intermediate results in the result set.

Step 1: Use a LEFT JOIN to combine Students, Subjects, and Examinations tables.
Step 2: Use COUNT() to aggregate the number of times each student attended each subject.
Step 3: Group the results by student_id and subject_name, and order them accordingly.

1. Students: [(1, Alice), (2, Bob), (13, John), (6, Alex)], Subjects: [(Math), (Science)], Examinations: [(1, Math), (1, Science), (2, Math)] 2. LEFT JOIN creates pairs for each student and subject. 3. For student 1 (Alice) and subject Math, count = 1 (found in Examinations) 4. For student 1 (Alice) and subject Science, count = 1 (found in Examinations) 5. For student 2 (Bob) and subject Math, count = 1 (found in Examinations) 6. For student 2 (Bob) and subject Science, count = 0 (not found).

Key Insights

Understanding how joins work in SQL is crucial for efficient data retrieval.
Using aggregation functions can simplify counting occurrences.

Common Mistakes

Not using GROUP BY correctly, leading to incorrect counts.
Forgetting to handle NULL values when using LEFT JOIN.

Interview Tips

Always clarify the requirements and constraints of the problem.
Practice writing SQL queries on paper to improve your fluency.
Understand the underlying data model and relationships between tables.

#1280

Students and Examinations

Easy

Database↗Hash Map↗Array↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

⚙️

Algorithm

3 steps

1Step 1: Use a LEFT JOIN to combine Students, Subjects, and Examinations tables.
2Step 2: Use COUNT() to aggregate the number of times each student attended each subject.
3Step 3: Group the results by student_id and subject_name, and order them accordingly.

solution.py7 lines

1# Full working Python code
2SELECT s.student_id, sub.subject_name, COUNT(e.student_id) AS attended
3FROM Students s
4CROSS JOIN Subjects sub
5LEFT JOIN Examinations e ON s.student_id = e.student_id AND sub.subject_name = e.subject_name
6GROUP BY s.student_id, sub.subject_name
7ORDER BY s.student_id, sub.subject_name;

ℹ

Complexity note: The time complexity is O(n) because we are performing a single pass through the data with joins and aggregations. The space complexity is O(n) due to the storage of intermediate results in the result set.

1Understanding how joins work in SQL is crucial for efficient data retrieval.
2Using aggregation functions can simplify counting occurrences.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.