How do you solve Number of Unique Subjects Taught by Each Teacher on LeetCode?

Number of Unique Subjects Taught by Each Teacher (LeetCode #2356) 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: Using a HashMap allows for efficient counting of unique items without needing nested loops.

What is the brute force solution for Number of Unique Subjects Taught by Each Teacher?

The brute force approach for Number of Unique Subjects Taught by Each Teacher is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each teacher and counting the unique subjects they teach by comparing against all other subjects. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Unique Subjects Taught by Each Teacher?

Number of Unique Subjects Taught by Each Teacher uses the following concepts: Database. The recommended patterns to study are: Hash Map, Set.

What are the interview tips for Number of Unique Subjects Taught by Each Teacher?

Always consider the efficiency of your approach; start with brute force but aim for optimization. Explain your thought process clearly; it shows your understanding of the problem. Practice writing code on a whiteboard or in a timed setting to simulate interview conditions.

What are common mistakes when solving Number of Unique Subjects Taught by Each Teacher?

Students often forget to handle duplicates when counting subjects. Not using efficient data structures like HashMap or Set can lead to performance issues.

#2356

Number of Unique Subjects Taught by Each Teacher

Easy

DatabaseHash MapSet

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)

The optimal solution uses a HashMap to efficiently count unique subjects for each teacher in a single pass through the data, significantly reducing the time complexity.

⚙️

Algorithm

5 steps

1Step 1: Initialize a HashMap to store each teacher's unique subjects.
2Step 2: Iterate through each row in the teacher table.
3Step 3: For each teacher_id, add the subject_id to the HashMap's set of subjects.
4Step 4: After processing all rows, create a result list from the HashMap.
5Step 5: Return the result list.

solution.py11 lines

1# Full working Python code
2import pandas as pd
3
4def count_unique_subjects(teacher_df):
5    subject_count = {}  
6    for _, row in teacher_df.iterrows():
7        if row['teacher_id'] not in subject_count:
8            subject_count[row['teacher_id']] = set()
9        subject_count[row['teacher_id']].add(row['subject_id'])
10    result = [{'teacher_id': teacher, 'cnt': len(subjects)} for teacher, subjects in subject_count.items()]
11    return pd.DataFrame(result)

ℹ

Complexity note: The time complexity is O(n) because we only go through the list of teachers once. The space complexity is O(n) due to the storage of unique subjects in the HashMap.

1Using a HashMap allows for efficient counting of unique items without needing nested loops.
2Sets are useful for automatically handling duplicates when counting unique subjects.

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