How do you solve Most Common Course Pairs on LeetCode?

Most Common Course Pairs (LeetCode #3764) 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: Focus on top performers to identify effective learning paths.

What is the time complexity of Most Common Course Pairs?

The optimal solution for Most Common Course Pairs runs in O(n) time and uses O(n) space. The optimal approach leverages a single pass through the data to count pairs, leading to linear time complexity.

What is the brute force solution for Most Common Course Pairs?

The brute force approach for Most Common Course Pairs is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible pair of courses for each top-performing student. This involves iterating through all course sequences and counting pairs, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Most Common Course Pairs?

Understand the problem requirements thoroughly. Practice writing SQL queries for data manipulation. Think about time and space complexity in your solutions.

What are common mistakes when solving Most Common Course Pairs?

Neglecting to filter for top-performing students. Forgetting to maintain order in course sequences.

#3764

Most Common Course Pairs

Hard

Hash MapArray

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)

We can use a hash map to efficiently count course pairs as we iterate through the course sequences of top-performing students, reducing the need for nested loops.

⚙️

Algorithm

3 steps

1Step 1: Filter top-performing students and store their course sequences.
2Step 2: Use a hash map to count occurrences of each course pair as we traverse each student's course list.
3Step 3: Return the pairs sorted by frequency.

solution.py1 lines

1SELECT course_a, course_b, COUNT(*) as pair_freq FROM (SELECT user_id, course_id as course_a, LEAD(course_id) OVER (PARTITION BY user_id ORDER BY completion_date) as course_b FROM course_completions WHERE user_id IN (SELECT user_id FROM course_completions GROUP BY user_id HAVING COUNT(*) >= 5 AND AVG(course_rating) >= 4)) AS pairs GROUP BY course_a, course_b ORDER BY pair_freq DESC;

ℹ

Complexity note: The optimal approach leverages a single pass through the data to count pairs, leading to linear time complexity.

1Focus on top performers to identify effective learning paths.
2Course sequences can reveal valuable insights into skill progression.

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