LeetCode 596: Classes More Than 5 Students Solution in Python – A Step-by-Step Guide

What Is LeetCode 596: Classes More Than 5 Students?

In LeetCode 596: Classes More Than 5 Students, you’re given a list of tuples courses where each tuple (student, class_name) represents a student enrolled in a class. Your task is to return a list of class names where the number of distinct students enrolled is strictly greater than 5. For example, with courses = [("Alice","Math"), ("Bob","Math"), ("Charlie","Math"), ("David","Math"), ("Eve","Math"), ("Frank","Math"), ("Grace","English")], the result is ["Math"] because "Math" has 6 students, while "English" has only 1. This problem builds on LeetCode 349: Intersection of Two Arrays for grouping but focuses on counting distinct elements with a threshold.

Problem Statement

• Input: courses (List[Tuple[str, str]])—list of (student, class_name) enrollments.
• Output: List[str]—class names with more than 5 distinct students.
• Rules: Count unique students per class; return classes with count > 5; students may enroll in multiple classes.

Constraints

• 1 <= courses.length <= 10⁴
• 1 <= student.length, class_name.length <= 50
• student and class_name consist of lowercase English letters

Examples

• Input: courses = [("Alice","Math"),("Bob","Math"),("Charlie","Math"),("David","Math"),("Eve","Math"),("Frank","Math"),("Grace","English")]
• Output: ["Math"]
• "Math": 6 students, "English": 1 student.
• Input: courses = [("Alice","Math"),("Bob","Math"),("Charlie","English")]
• Output: []
• "Math": 2 students, "English": 1 student, none > 5.
• Input: courses = [("Alice","Math"),("Alice","Math"),("Bob","Math"),("Charlie","Math"),("David","Math"),("Eve","Math"),("Frank","Math")]
• Output: ["Math"]
• "Math": 6 unique students (Alice counted once).

Understanding the Problem: Counting Crowded Classes

To solve LeetCode 596: Classes More Than 5 Students in Python, we need to count the number of distinct students per class in a list of enrollments and return class names with more than 5 students, handling up to 10⁴ records efficiently. A brute-force approach scanning the list repeatedly for each class could work but is inefficient (O(n²)). Instead, a dictionary-based counting method with sets processes the list in O(n) time, grouping students by class and counting uniques using set operations. We’ll explore:

• Best Solution (Dictionary-Based Counting with Filtering): O(n) time, O(n) space—fast and optimal (n = number of enrollments).
• Alternative Solution (Brute-Force Traversal): O(n²) time, O(n) space—thorough but slower.

Let’s dive into the dictionary solution with a friendly breakdown!

Best Solution: Dictionary-Based Counting with Filtering

Why Dictionary Wins

The dictionary-based counting with filtering solution is the best for LeetCode 596 because it identifies classes with more than 5 students in O(n) time and O(n) space by using a dictionary to map each class to a set of unique student names in a single pass, then filtering classes based on set size efficiently. It’s like keeping a class roster, jotting down each student’s name once per class, and spotlighting the crowded ones—all in a quick, organized tally!

How It Works

Think of this as a class-enrollment organizer:

• Step 1: Build Dictionary with Sets:
• Map each class_name to a set of student names from courses.
• Step 2: Count and Filter:
• Iterate dictionary, check if set size > 5, collect class names.
• Step 3: Return Result:
• List of class names with > 5 students.
• Why It Works:
• Sets ensure distinct student counts.
• Single pass builds the map efficiently.

It’s like a crowded-class spotting maestro!

Step-by-Step Example

Example: courses = [("Alice","Math"),("Bob","Math"),("Charlie","Math"),("David","Math"),("Eve","Math"),("Frank","Math"),("Grace","English")]

• Step 1: Build dictionary:
• class_students = {"Math": {"Alice", "Bob", "Charlie", "David", "Eve", "Frank"}, "English": {"Grace"}}.
• Step 2: Count and filter:
• "Math": size = 6 > 5, include "Math".
• "English": size = 1 < 5, skip.
• Step 3: Result: ["Math"].
• Result: ["Math"].

Example: courses = [("Alice","Math"),("Bob","Math"),("Charlie","English")]

• Step 1: class_students = {"Math": {"Alice", "Bob"}, "English": {"Charlie"}}.
• Step 2:
• "Math": size = 2 < 5, skip.
• "English": size = 1 < 5, skip.
• Step 3: Result: [].
• Result: [].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codefrom typing import List, Tuple
from collections import defaultdict

class Solution:
    def findClasses(self, courses: List[Tuple[str, str]]) -> List[str]:
        # Step 1: Build dictionary with sets of students per class
        class_students = defaultdict(set)
        for student, class_name in courses:
            class_students[class_name].add(student)

        # Step 2: Filter classes with more than 5 students
        result = [class_name for class_name, students in class_students.items() 
                  if len(students) > 5]

        # Step 3: Return result
        return result

• Line 7-9: Use defaultdict(set) to map classes to student sets, add each student.
• Line 12-13: List comprehension:
• Iterate dictionary, check set size > 5, collect class names.
• Line 16: Return list of qualifying classes.
• Time Complexity: O(n)—single pass to build map and filter.
• Space Complexity: O(n)—dictionary and sets storage.

It’s like a class-counting wizard!

Alternative Solution: Brute-Force Traversal

Why an Alternative Approach?

The brute-force traversal approach scans the list repeatedly to count unique students per class, running in O(n²) time and O(n) space for storing unique students per class. It’s thorough but inefficient, making it a good baseline for understanding or when avoiding advanced data structures.

How It Works

Picture this as a class-scanning seeker:

• Step 1: Get unique class names.
• Step 2: For each class, scan list to count unique students.
• Step 3: Add classes with > 5 students to result.
• Step 4: Return result list.

It’s like a manual class counter!

Step-by-Step Example

Example: courses = [("Alice","Math"),("Bob","Math"),("Charlie","English")]

• Step 1: Unique classes: ["Math", "English"].
• Step 2: Count:
• "Math": Scan → {"Alice", "Bob"}, size = 2.
• "English": Scan → {"Charlie"}, size = 1.
• Step 3: No class > 5, result = [].
• Result: [].

Code for Brute-Force Approach

Copy codefrom typing import List, Tuple

class Solution:
    def findClasses(self, courses: List[Tuple[str, str]]) -> List[str]:
        # Step 1: Get unique class names
        classes = set(class_name for _, class_name in courses)

        # Step 2: Count unique students per class
        result = []
        for class_name in classes:
            students = set()
            for student, enrolled_class in courses:
                if enrolled_class == class_name:
                    students.add(student)
            if len(students) > 5:
                result.append(class_name)

        # Step 3: Return result
        return result

• Line 6: Extract unique class names.
• Lines 9-15: For each class, build set of students, check size > 5.
• Line 18: Return list.
• Time Complexity: O(n²)—nested loops.
• Space Complexity: O(n)—sets for students.

It’s a brute-force class tallier!

Comparing the Two Solutions

• Dictionary (Best):
• Pros: O(n), O(n), fast and elegant.
• Cons: Extra space for dictionary.
• Brute-Force (Alternative):
• Pros: O(n²), O(n), explicit counting.
• Cons: Slower, less efficient.

Dictionary wins for efficiency!

Additional Examples and Edge Cases

• Empty list: [].
• Duplicates: Counted once per class.
• Small classes: [].

Dictionary handles them all!

Complexity Recap

• Dictionary: Time O(n), Space O(n).
• Brute-Force: Time O(n²), Space O(n).

Dictionary’s the speed champ!

Key Takeaways

• Dictionary: Counting finesse—learn at Python Basics!
• Brute-Force: Scan simplicity.
• Classes: Crowds are fun.
• Python: Dict or loops, your pick!

Final Thoughts: Spot Those Crowded Classes!

LeetCode 596: Classes More Than 5 Students in Python is a practical data challenge. Dictionary-based counting with filtering is your fast track, while brute-force offers a thorough dive. Want more? Try LeetCode 349: Intersection of Two Arrays or LeetCode 560: Subarray Sum Equals K. Ready to count? Head to Solve LeetCode 596 on LeetCode and identify those crowded classes today!