LeetCode 580: Count Student Number in Departments Solution in Python – A Step-by-Step Guide

What Is LeetCode 580: Count Student Number in Departments?

In LeetCode 580: Count Student Number in Departments, you’re given two lists: departments, a list of tuples (dept_id, dept_name) representing department ID and name, and students, a list of tuples (student_id, dept_id) representing student ID and the department they’re enrolled in. Your task is to return a dictionary mapping each dept_name to the number of students enrolled in that department, including departments with zero students. For example, with departments = [(1, "CS"), (2, "Math")] and students = [(101, 1), (102, 1), (103, 2)], the result is {"CS": 2, "Math": 1}. This problem builds on LeetCode 349: Intersection of Two Arrays for joining data but focuses on counting within groups.

Problem Statement

• Input: departments (List[Tuple[int, str]])—list of (dept_id, dept_name); students (List[Tuple[int, int]])—list of (student_id, dept_id).
• Output: Dict[str, int]—mapping of dept_name to student count (including 0 counts).
• Rules: Match dept_id between lists; count students per department; include all departments.

Constraints

• 1 <= departments.length, students.length <= 10⁵
• 1 <= dept_id, student_id <= 10⁵ (unique in departments)
• 1 <= dept_name.length <= 10

Examples

• Input: departments = [(1,"CS"),(2,"Math")], students = [(101,1),(102,1),(103,2)]
• Output: {"CS": 2, "Math": 1}
• CS has 2 students, Math has 1.
• Input: departments = [(1,"CS"),(2,"Math")], students = [(101,1)]
• Output: {"CS": 1, "Math": 0}
• CS has 1 student, Math has 0.
• Input: departments = [(1,"CS")], students = []
• Output: {"CS": 0}
• No students enrolled.

Understanding the Problem: Counting Students

To solve LeetCode 580: Count Student Number in Departments in Python, we need to join the departments and students lists by dept_id, count the number of students per department, and return a dictionary mapping each dept_name to its student count, ensuring all departments are included (even with zero students), handling up to 10⁵ records efficiently. A brute-force nested loop (O(n * m)) could work but would be slow for large lists. Instead, a dictionary-based grouping preprocesses data for O(n + m) performance (n = departments, m = students), making it fast and scalable. We’ll explore:

• Best Solution (Dictionary-Based Grouping): O(n + m) time, O(n) space—fast and optimal.
• Alternative Solution (Brute-Force Traversal): O(n * m) time, O(1) space—simple but slow.

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

Best Solution: Dictionary-Based Grouping

Why Dictionary Wins

The dictionary-based grouping solution is the best for LeetCode 580 because it efficiently counts students per department in O(n + m) time and O(n) space by using two dictionaries: one to map department IDs to names and another to tally student counts, processing each list in a single pass, avoiding redundant searches. It’s like keeping a department roster and a student counter, quickly tallying enrollments as they come in—all in a smooth, streamlined process!

How It Works

Think of this as a student-enrollment organizer:

• Step 1: Build Department Dictionary:
• Map dept_id to dept_name from departments.
• Step 2: Count Students:
• Use a dictionary to tally students per dept_id from students.
• Step 3: Initialize Result:
• Map all dept_name to 0 in result dictionary.
• Step 4: Update Counts:
• Transfer student counts to result using department names.
• Step 5: Return Result:
• Dictionary with student counts per department.
• Why It Works:
• Dictionaries provide O(1) lookups.
• Single pass per list ensures efficiency.

It’s like a student-counting maestro!

Step-by-Step Example

Example: departments = [(1,"CS"),(2,"Math")], students = [(101,1),(102,1),(103,2)]

• Step 1: Build dictionary:
• id_to_name = {1: "CS", 2: "Math"}.
• Step 2: Count students:
• (101, 1): counts = {1: 1}.
• (102, 1): counts = {1: 2}.
• (103, 2): counts = {1: 2, 2: 1}.
• Step 3: Initialize result:
• result = {"CS": 0, "Math": 0}.
• Step 4: Update counts:
• result["CS"] = 2, result["Math"] = 1.
• Result: {"CS": 2, "Math": 1}.

Example: departments = [(1,"CS"),(2,"Math")], students = [(101,1)]

• Step 1: id_to_name = {1: "CS", 2: "Math"}.
• Step 2: counts = {1: 1}.
• Step 3: result = {"CS": 0, "Math": 0}.
• Step 4: result["CS"] = 1.
• Result: {"CS": 1, "Math": 0}.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codefrom typing import List, Tuple

class Solution:
    def countStudents(self, departments: List[Tuple[int, str]], students: List[Tuple[int, int]]) -> dict[str, int]:
        # Step 1: Build dictionary mapping dept_id to dept_name
        id_to_name = {dept_id: dept_name for dept_id, dept_name in departments}

        # Step 2: Count students per department
        dept_counts = {}
        for _, dept_id in students:
            dept_counts[dept_id] = dept_counts.get(dept_id, 0) + 1

        # Step 3: Initialize result with all departments at 0
        result = {dept_name: 0 for _, dept_name in departments}

        # Step 4: Update result with student counts
        for dept_id, count in dept_counts.items():
            if dept_id in id_to_name:
                result[id_to_name[dept_id]] = count

        # Step 5: Return result dictionary
        return result

• Line 6: Create dictionary from departments using dict comprehension.
• Lines 9-11: Tally student counts per dept_id.
• Line 14: Initialize result with all department names set to 0.
• Lines 17-19: Update result with counts where dept_id matches.
• Line 22: Return dictionary.
• Time Complexity: O(n + m)—one pass each for departments and students.
• Space Complexity: O(n)—dictionary storage (n = departments).

It’s like a student-tallying wizard!

Alternative Solution: Brute-Force Traversal

Why an Alternative Approach?

The brute-force traversal approach iterates each department and scans the entire student list to count matches, running in O(n * m) time and O(1) space (excluding output). It’s simple but inefficient for large lists, making it a good baseline for small datasets or understanding.

How It Works

Picture this as a student-scanning seeker:

• Step 1: Iterate each department.
• Step 2: Scan students for matching dept_id, count matches.
• Step 3: Build result dictionary with counts.
• Step 4: Return result.

It’s like a student-counting explorer!

Step-by-Step Example

Example: departments = [(1,"CS"),(2,"Math")], students = [(101,1)]

• Step 1: Dept (1, "CS"):
• Scan: (101, 1) matches, count = 1.
• Step 2: Dept (2, "Math"):
• Scan: No matches, count = 0.
• Step 3: Result: {"CS": 1, "Math": 0}.
• Result: {"CS": 1, "Math": 0}.

Code for Brute-Force Approach

Copy codefrom typing import List, Tuple

class Solution:
    def countStudents(self, departments: List[Tuple[int, str]], students: List[Tuple[int, int]]) -> dict[str, int]:
        # Step 1: Initialize result dictionary
        result = {}

        # Step 2: Count students for each department
        for dept_id, dept_name in departments:
            count = 0
            for _, student_dept_id in students:
                if student_dept_id == dept_id:
                    count += 1
            result[dept_name] = count

        # Step 3: Return result
        return result

• Line 7: Empty dict for results.
• Lines 10-14: For each department, count matching students.
• Line 17: Return dictionary.
• Time Complexity: O(n * m)—nested loops.
• Space Complexity: O(1)—minimal extra space (excluding output).

It’s a brute-force student counter!

Comparing the Two Solutions

• Dictionary (Best):
• Pros: O(n + m), O(n), fast.
• Cons: Extra dictionary space.
• Brute-Force (Alternative):
• Pros: O(n * m), O(1), space-efficient.
• Cons: Slower.

Dictionary wins for efficiency!

Additional Examples and Edge Cases

• Empty students: All counts 0.
• No matches: Some counts 0.
• Single dept: Single count.

Dictionary handles them all!

Complexity Recap

• Dictionary: Time O(n + m), Space O(n).
• Brute-Force: Time O(n * m), Space O(1).

Dictionary’s the speed champ!

Key Takeaways

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

Final Thoughts: Tally Those Students!

LeetCode 580: Count Student Number in Departments in Python is a practical data challenge. Dictionary-based grouping 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 580 on LeetCode and tally those students today!