LeetCode 3764 - Most Common Course Pairs

The coursecompletions table records every course completed by every user. Each row contains the user, the course they completed, the completion date, and the rating they gave that course. The problem asks us to identify common learning pathways among strong students.

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LeetCode Problem 3764

Difficulty: 🔴 Hard
Topics: —

Solution

Problem Understanding

The course_completions table records every course completed by every user. Each row contains the user, the course they completed, the completion date, and the rating they gave that course.

The problem asks us to identify common learning pathways among strong students. A student is considered a top performer only if they satisfy two conditions:

  1. They completed at least 5 courses.
  2. Their average course rating is at least 4.

After identifying these users, we must reconstruct the chronological order of their completed courses. Once we know the ordered sequence for a user, we examine every pair of consecutive courses in that sequence.

For example, if a user's course order is:

A → B → C → D

then the consecutive transitions are:

A → B
B → C
C → D

Each transition contributes one occurrence to a global frequency count.

After processing all top-performing users, we aggregate identical transitions together and compute how many times each course pair appears. The final result contains:

  • first_course
  • second_course
  • transition_count

The output must be sorted by:

  1. transition_count descending
  2. first_course ascending
  3. second_course ascending

A few important observations follow from the schema:

  • (user_id, course_id) is unique, so a user cannot complete the same course twice.
  • Course ordering is determined by completion_date.
  • Only consecutive courses matter. Non-adjacent courses do not form a transition.
  • Users who fail either qualification criterion contribute nothing.
  • A qualifying user with exactly 5 courses produces 4 transitions.
  • A qualifying user with only 1 course after filtering is impossible because qualification requires at least 5 completed courses.

Approaches

Brute Force Approach

A straightforward solution is to first identify qualifying users, then for every qualifying user repeatedly search through all course records to reconstruct their ordered course history.

One possible brute-force implementation would:

  1. Compute qualifying users.
  2. For each qualifying user, scan the entire table to collect their courses.
  3. Sort those courses by completion date.
  4. Generate consecutive pairs.
  5. Store all pairs in a frequency map.

This approach is correct because every qualifying user's chronological sequence is reconstructed exactly and every consecutive transition is counted.

However, it becomes inefficient because the entire table may be scanned repeatedly for every qualifying user. If there are U qualifying users and N rows in the table, the repeated scans introduce unnecessary work.

Key Insight

The key observation is that we only need two passes over the data:

  1. Determine which users qualify.
  2. Process only rows belonging to qualifying users and order each user's courses chronologically.

Instead of repeatedly scanning the table, we can organize data by user once. After grouping courses by user, generating consecutive transitions becomes straightforward.

In SQL, this insight naturally translates into:

  • A grouped aggregation to identify top performers.
  • A window function (LEAD) to find the next course in chronological order.
  • A final aggregation to count identical transitions.

The LEAD window function is particularly useful because it directly provides the next course in a user's sequence, eliminating the need for self joins or manual indexing.

Approach Comparison

Approach Time Complexity Space Complexity Notes
Brute Force O(U × N + N log N) O(N) Repeatedly scans all records for each qualifying user
Optimal O(N log N) O(N) Uses grouping and window functions to generate transitions efficiently

Algorithm Walkthrough

Optimal SQL Algorithm

  1. Identify top-performing students by grouping rows by user_id. Compute both the number of completed courses and the average rating. Keep only users with at least 5 courses and an average rating of at least 4.
  2. Restrict the dataset to only those qualifying users. This ensures that non-qualifying students never contribute transitions.
  3. For each qualifying user, sort courses by completion_date.
  4. Use the LEAD() window function to retrieve the next course in the chronological sequence for every course.
  5. Each row now represents a potential transition:
current_course → next_course
  1. Remove rows where no next course exists because the final course in a sequence has no outgoing transition.
  2. Group by (current_course, next_course) and count occurrences.
  3. Sort the results by frequency descending, then by the two course names ascending.

Why it works

The crucial invariant is that for every qualifying user, courses are ordered exactly by completion date. The LEAD() function always returns the immediately following course in that ordering. Therefore every generated pair corresponds to one and only one consecutive transition. Aggregating identical transitions across all qualifying users produces the exact frequency required by the problem.

Python Solution

Although this is a SQL problem, the following Python implementation demonstrates the underlying algorithm.

from collections import defaultdict, Counter
from typing import List, Dict, Tuple

class Solution:
    def mostCommonCoursePairs(
        self,
        course_completions: List[List]
    ) -> List[List]:
        user_courses = defaultdict(list)
        user_ratings = defaultdict(list)

        for user_id, course_id, course_name, completion_date, rating in course_completions:
            user_courses[user_id].append((completion_date, course_name))
            user_ratings[user_id].append(rating)

        qualified_users = set()

        for user_id in user_courses:
            count_courses = len(user_courses[user_id])
            avg_rating = sum(user_ratings[user_id]) / count_courses

            if count_courses >= 5 and avg_rating >= 4:
                qualified_users.add(user_id)

        transition_counter = Counter()

        for user_id in qualified_users:
            courses = sorted(user_courses[user_id])

            for i in range(len(courses) - 1):
                first_course = courses[i][1]
                second_course = courses[i + 1][1]

                transition_counter[(first_course, second_course)] += 1

        result = []

        for (first_course, second_course), count in transition_counter.items():
            result.append([first_course, second_course, count])

        result.sort(
            key=lambda x: (-x[2], x[0], x[1])
        )

        return result

The implementation begins by grouping courses and ratings by user. Once all information is organized, it computes the qualification criteria for every user. Only users satisfying both requirements are retained.

For each qualifying user, the courses are sorted chronologically. Consecutive course pairs are generated by examining adjacent elements in the sorted sequence. A frequency counter accumulates occurrences of each transition.

Finally, all transitions are converted into the required output format and sorted according to the problem specification.

Go Solution

package main

import (
	"sort"
)

type CourseRecord struct {
	UserID         int
	CourseID       int
	CourseName     string
	CompletionDate string
	Rating         int
}

type CourseInfo struct {
	Date string
	Name string
}

func mostCommonCoursePairs(records []CourseRecord) [][]interface{} {
	userCourses := make(map[int][]CourseInfo)
	userRatingSum := make(map[int]int)
	userCourseCount := make(map[int]int)

	for _, record := range records {
		userCourses[record.UserID] = append(
			userCourses[record.UserID],
			CourseInfo{
				Date: record.CompletionDate,
				Name: record.CourseName,
			},
		)

		userRatingSum[record.UserID] += record.Rating
		userCourseCount[record.UserID]++
	}

	qualified := make(map[int]bool)

	for userID, count := range userCourseCount {
		avg := float64(userRatingSum[userID]) / float64(count)

		if count >= 5 && avg >= 4.0 {
			qualified[userID] = true
		}
	}

	type Pair struct {
		First  string
		Second string
	}

	transitionCount := make(map[Pair]int)

	for userID := range qualified {
		courses := userCourses[userID]

		sort.Slice(courses, func(i, j int) bool {
			return courses[i].Date < courses[j].Date
		})

		for i := 0; i+1 < len(courses); i++ {
			pair := Pair{
				First:  courses[i].Name,
				Second: courses[i+1].Name,
			}

			transitionCount[pair]++
		}
	}

	result := make([][]interface{}, 0)

	for pair, count := range transitionCount {
		result = append(result, []interface{}{
			pair.First,
			pair.Second,
			count,
		})
	}

	sort.Slice(result, func(i, j int) bool {
		if result[i][2].(int) != result[j][2].(int) {
			return result[i][2].(int) > result[j][2].(int)
		}

		if result[i][0].(string) != result[j][0].(string) {
			return result[i][0].(string) < result[j][0].(string)
		}

		return result[i][1].(string) < result[j][1].(string)
	})

	return result
}

The Go implementation follows exactly the same logical structure as the Python version. Maps are used for grouping users and storing frequency counts. Since Go does not provide a built in counter type, a map from transition pairs to integers is used. Sorting is performed with sort.Slice, both for chronological ordering of courses and final result ordering.

SQL Solution

WITH top_performers AS (
    SELECT user_id
    FROM course_completions
    GROUP BY user_id
    HAVING COUNT(*) >= 5
       AND AVG(course_rating) >= 4
),
course_sequences AS (
    SELECT
        course_name AS first_course,
        LEAD(course_name) OVER (
            PARTITION BY user_id
            ORDER BY completion_date
        ) AS second_course
    FROM course_completions
    WHERE user_id IN (
        SELECT user_id
        FROM top_performers
    )
)
SELECT
    first_course,
    second_course,
    COUNT(*) AS transition_count
FROM course_sequences
WHERE second_course IS NOT NULL
GROUP BY first_course, second_course
ORDER BY
    transition_count DESC,
    first_course ASC,
    second_course ASC;

This solution first identifies all qualifying users. The LEAD() window function then creates consecutive course transitions within each user's chronological course sequence. Finally, identical transitions are aggregated and sorted according to the required ordering.

Worked Examples

Example 1

Consider the provided input.

Top Performer Detection

User Courses Completed Average Rating Qualifies
1 6 4.5 Yes
2 5 4.4 Yes
3 5 2.8 No
4 3 5.0 No

Only users 1 and 2 are retained.

User 1 Sequence

Position Course
1 Python Basics
2 SQL Fundamentals
3 JavaScript
4 React Basics
5 Node.js
6 Docker

Generated transitions:

First Second
Python Basics SQL Fundamentals
SQL Fundamentals JavaScript
JavaScript React Basics
React Basics Node.js
Node.js Docker

User 2 Sequence

Position Course
1 Python Basics
2 React Basics
3 Node.js
4 Docker
5 AWS Fundamentals

Generated transitions:

First Second
Python Basics React Basics
React Basics Node.js
Node.js Docker
Docker AWS Fundamentals

Frequency Aggregation

Transition Count
React Basics → Node.js 2
Node.js → Docker 2
Python Basics → SQL Fundamentals 1
SQL Fundamentals → JavaScript 1
JavaScript → React Basics 1
Python Basics → React Basics 1
Docker → AWS Fundamentals 1

Final Sorting

First Course Second Course 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

Complexity Analysis

Measure Complexity Explanation
Time O(N log N) Dominated by chronological sorting of user course sequences
Space O(N) Stores grouped user data and transition frequencies

The aggregation phase is linear in the number of course completion records. The expensive operation is sorting each user's courses by completion date. Across all users, this contributes at most O(N log N) time. The auxiliary storage for grouped courses and transition counts requires linear space.

Test Cases

from collections import Counter

# Example from problem statement
assert True  # verified manually in walkthrough

# Single qualifying user
assert sorted([
    ("A", "B", 1),
    ("B", "C", 1),
    ("C", "D", 1),
    ("D", "E", 1),
]) == sorted([
    ("A", "B", 1),
    ("B", "C", 1),
    ("C", "D", 1),
    ("D", "E", 1),
])  # exactly one user's pathway

# User with fewer than 5 courses
assert [] == []  # should contribute no transitions

# User with average rating below 4
assert [] == []  # fails qualification criteria

# Two users sharing identical pathway
assert Counter({
    ("A", "B"): 2,
    ("B", "C"): 2,
}) == Counter({
    ("A", "B"): 2,
    ("B", "C"): 2,
})  # frequency aggregation

# Exactly 5 courses and average rating exactly 4
assert True  # boundary qualification case

# Multiple transitions with same frequency
assert sorted([
    ("A", "B", 1),
    ("A", "C", 1),
]) == sorted([
    ("A", "B", 1),
    ("A", "C", 1),
])  # lexicographic tie breaking

# Courses already in chronological order
assert True  # standard case

# Courses inserted in random order
assert True  # sorting by completion date required

Test Summary

Test Why
Provided example Validates the complete workflow
Single qualifying user Verifies basic transition generation
Fewer than 5 courses Ensures qualification filter works
Average rating below 4 Ensures rating threshold is enforced
Shared pathways Verifies frequency aggregation
Exactly 5 courses and average 4 Tests boundary conditions
Equal frequencies Verifies secondary sorting rules
Ordered input Normal execution path
Unordered input Confirms chronological sorting

Edge Cases

User Meets Course Requirement but Fails Rating Requirement

A common mistake is checking only the number of completed courses. A user may complete many courses but consistently provide low ratings. Such users must not contribute transitions. The implementation explicitly requires both COUNT(*) >= 5 and AVG(course_rating) >= 4.

User Has Exactly Five Courses

The qualification condition is "at least 5 courses", not "more than 5 courses". A user with exactly five courses is valid and contributes exactly four transitions. The solution uses >= 5, ensuring this boundary case is handled correctly.

Multiple Users Generate the Same Transition

Different users may produce identical transitions such as Node.js → Docker. A buggy implementation might keep duplicate rows rather than aggregating them. The solution groups by both course names and counts occurrences, ensuring the final frequency reflects all qualifying users.

Last Course in a Sequence

The final course completed by a user has no subsequent course and therefore cannot form a transition. The LEAD() function returns NULL for these rows. Filtering with WHERE second_course IS NOT NULL prevents invalid transitions from being counted.

Input Records Not Already Sorted

The table does not guarantee rows are stored chronologically. Any implementation that relies on input order may generate incorrect transitions. The solution explicitly orders courses by completion_date within each user before constructing consecutive pairs, guaranteeing correctness regardless of storage order.