Skip to main content
Back to problems
#3764
Hard Database

Most common course pairs

51.1% acceptance
Feb 27, 2026
27
1
Table: course_completions +-------------------+---------+ | Column Name | Type | +-------------------+---------+ | user_id | int | | course_id | int | | course_name | varchar | | completion_date | date | | course_rating | int | +-------------------+---------+ (user_id, course_id) is the combination of columns with unique values for this table. Each row represents a completed course by a user with their rating (1-5 scale). Write a solution to identify skill mastery pathways by analyzing course completion sequences among top-performing students: Consider only top-performing students (those who completed at least 5 courses with an average rating of 4 or higher). For each top performer, identify the sequence of courses they completed in chronological order. Find all consecutive course pairs (Course A → Course B) taken by these students. Return the pair frequency, identifying which course transitions are most common among high achievers. Return the result table ordered by pair frequency in descending order and then by first course name and second course name in ascending order. The result format is in the following example.

Solution

SQL
LeetCode
solution.sql
#
# Table: course_completions
# +-------------------+---------+
# | Column Name       | Type    |
# +-------------------+---------+
# | user_id           | int     |
# | course_id         | int     |
# | course_name       | varchar |
# | completion_date   | date    |
# | course_rating     | int     |
# +-------------------+---------+
# (user_id, course_id) is the combination of columns with unique values for this table.
# Each row represents a completed course by a user with their rating (1-5 scale).
# Write a solution to identify skill mastery pathways by analyzing course completion sequences among top-performing students:
# Consider only top-performing students (those who completed at least 5 courses with an average rating of 4 or higher).
# For each top performer, identify the sequence of courses they completed in chronological order.
# Find all consecutive course pairs (Course A → Course B) taken by these students.
# Return the pair frequency, identifying which course transitions are most common among high achievers.
# Return the result table ordered by pair frequency in descending order and then by first course name and second course name in ascending order.
# The result format is in the following example.
# Example:
# Input:
# course_completions table:
# +---------+-----------+------------------+-----------------+---------------+
# | user_id | course_id | course_name      | completion_date | course_rating |
# +---------+-----------+------------------+-----------------+---------------+
# | 1       | 101       | Python Basics    | 2024-01-05      | 5             |
# | 1       | 102       | SQL Fundamentals | 2024-02-10      | 4             |
# | 1       | 103       | JavaScript       | 2024-03-15      | 5             |
# | 1       | 104       | React Basics     | 2024-04-20      | 4             |
# | 1       | 105       | Node.js          | 2024-05-25      | 5             |
# | 1       | 106       | Docker           | 2024-06-30      | 4             |
# | 2       | 101       | Python Basics    | 2024-01-08      | 4             |
# | 2       | 104       | React Basics     | 2024-02-14      | 5             |
# | 2       | 105       | Node.js          | 2024-03-20      | 4             |
# | 2       | 106       | Docker           | 2024-04-25      | 5             |
# | 2       | 107       | AWS Fundamentals | 2024-05-30      | 4             |
# | 3       | 101       | Python Basics    | 2024-01-10      | 3             |
# | 3       | 102       | SQL Fundamentals | 2024-02-12      | 3             |
# | 3       | 103       | JavaScript       | 2024-03-18      | 3             |
# | 3       | 104       | React Basics     | 2024-04-22      | 2             |
# | 3       | 105       | Node.js          | 2024-05-28      | 3             |
# | 4       | 101       | Python Basics    | 2024-01-12      | 5             |
# | 4       | 108       | Data Science     | 2024-02-16      | 5             |
# | 4       | 109       | Machine Learning | 2024-03-22      | 5             |
# +---------+-----------+------------------+-----------------+---------------+
# Output:
# +------------------+------------------+------------------+
# | first_course     | second_course    | transition_count |
# +------------------+------------------+------------------+
# | Node.js          | Docker           | 2                |
# | React Basics     | Node.js          | 2                |
# | Docker           | AWS Fundamentals | 1                |
# | JavaScript       | React Basics     | 1                |
# | Python Basics    | React Basics     | 1                |
# | Python Basics    | SQL Fundamentals | 1                |
# | SQL Fundamentals | JavaScript       | 1                |
# +------------------+------------------+------------------+
# Explanation:
# User 1: Completed 6 courses with average rating 4.5 (qualifies as top performer)
# User 2: Completed 5 courses with average rating 4.4 (qualifies as top performer)
# User 3: Completed 5 courses but average rating is 2.8 (does not qualify)
# User 4: Completed only 3 courses (does not qualify)
# Course Pairs Among Top Performers:
# User 1: Python Basics → SQL Fundamentals → JavaScript → React Basics → Node.js → Docker
# User 2: Python Basics → React Basics → Node.js → Docker → AWS Fundamentals
# Most common transitions: Node.js → Docker (2 times), React Basics → Node.js (2 times)
# Results are ordered by transition_count in descending order, then by first_course in ascending order, and then by second_course in ascending order.
#

# Write your MySQL query statement below

WITH top_performers AS (
  SELECT user_id
  FROM course_completions
  GROUP BY user_id
  HAVING COUNT(*) >= 5 AND AVG(course_rating) >= 4
),
ordered_courses AS (
  SELECT cc.user_id, cc.course_name, cc.completion_date,
    ROW_NUMBER() OVER (PARTITION BY cc.user_id ORDER BY cc.completion_date) AS rn
  FROM course_completions cc
  JOIN top_performers tp ON cc.user_id = tp.user_id
),
pairs AS (
  SELECT a.course_name AS first_course, b.course_name AS second_course
  FROM ordered_courses a
  JOIN ordered_courses b ON a.user_id = b.user_id AND a.rn + 1 = b.rn
)
SELECT first_course, second_course, COUNT(*) AS transition_count
FROM pairs
GROUP BY first_course, second_course
ORDER BY transition_count DESC, first_course ASC, second_course ASC;