Is Maximum Students on a Single Bench easy or hard?

Maximum Students on a Single Bench is typically classified as an Easy problem. The solution relies on a basic frequency-count technique with a hash table and a single pass through the array.

Maximum Students on a Single Bench Python/Java solution

In Python, use a dictionary to track frequencies while iterating through the array. In Java, use a HashMap to store and update counts. Both implementations run in O(n) time and O(n) space using the same frequency-count pattern.

How to solve Maximum Students on a Single Bench in O(n)?

Use a hash map where the key is the bench number and the value is the number of students sitting there. Iterate through the array, increment the count for the current bench, and update the maximum frequency seen so far. This single-pass counting approach achieves O(n) time complexity.

What is the best approach for Maximum Students on a Single Bench?

The best approach uses a hash table to count how many times each bench number appears. Iterate through the array once, update the frequency in a map, and track the maximum count. This method runs in O(n) time with O(n) space and is the standard interview solution.

Is Maximum Students on a Single Bench asked at Google/Amazon/Meta?

Frequency counting problems using hash tables appear frequently in interviews at companies like Google, Amazon, and Meta. While this exact question may vary, the underlying pattern—tracking element frequencies efficiently—is a common interview topic.

What data structure is used in Maximum Students on a Single Bench?

The core data structure is a hash table (or hash map). It stores each bench number as a key and the number of students sitting on that bench as the value, enabling constant-time updates and lookups during iteration.

What is the time complexity of Maximum Students on a Single Bench?

The optimal solution runs in O(n) time because the array is scanned once and each hash map operation is O(1) on average. A naive brute force approach that compares each student with every other student takes O(n^2) time.

3450. Maximum Students on a Single Bench - Practice Online

You are given a 2D integer array of student data students, where students[i] = [student_id, bench_id] represents that student student_id is sitting on the bench bench_id.

Return the maximum number of unique students sitting on any single bench. If no students are present, return 0.

Note: A student can appear multiple times on the same bench in the input, but they should be counted only once per bench.

Example 1:

Input: students = [[1,2],[2,2],[3,3],[1,3],[2,3]]

Output: 3

Explanation:

Bench 2 has two unique students: [1, 2].
Bench 3 has three unique students: [1, 2, 3].
The maximum number of unique students on a single bench is 3.

Example 2:

Input: students = [[1,1],[2,1],[3,1],[4,2],[5,2]]

Output: 3

Explanation:

Bench 1 has three unique students: [1, 2, 3].
Bench 2 has two unique students: [4, 5].
The maximum number of unique students on a single bench is 3.

Example 3:

Input: students = [[1,1],[1,1]]

Output: 1

Explanation:

The maximum number of unique students on a single bench is 1.

Example 4:

Input: students = []

Output: 0

Explanation:

Since no students are present, the output is 0.

Constraints:

0 <= students.length <= 100
students[i] = [student_id, bench_id]
1 <= student_id <= 100
1 <= bench_id <= 100