How do you solve Maximum Number of Groups Entering a Competition on LeetCode?

Maximum Number of Groups Entering a Competition (LeetCode #2358) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting the grades helps in easily forming valid groups.

What is the time complexity of Maximum Number of Groups Entering a Competition?

The optimal solution for Maximum Number of Groups Entering a Competition runs in O(n log n) time and uses O(1) space. The sorting step takes O(n log n), and the subsequent iteration through the grades is O(n), making this approach efficient overall.

What is the brute force solution for Maximum Number of Groups Entering a Competition?

The brute force approach for Maximum Number of Groups Entering a Competition is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries all possible combinations of groups to find the maximum number of valid groups. This involves checking every possible way to partition the grades into groups, which can be quite slow. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Number of Groups Entering a Competition?

Maximum Number of Groups Entering a Competition uses the following concepts: Array, Math, Binary Search, Greedy. The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Maximum Number of Groups Entering a Competition?

Always consider edge cases, such as when all grades are the same. Explain your thought process clearly while coding, as it helps interviewers understand your approach. Practice similar problems to recognize patterns in group formation.

What are common mistakes when solving Maximum Number of Groups Entering a Competition?

Not sorting the grades before attempting to group them. Failing to reset the sum for the next group after forming a valid group.

#2358

Maximum Number of Groups Entering a Competition

Medium

ArrayMathBinary SearchGreedyGreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

💡

Intuition

Time O(n log n)Space O(1)

The optimal solution leverages the sorted order of grades to systematically form groups of increasing size and sum. By sorting the grades first, we can easily ensure that the conditions for group formation are met.

⚙️

Algorithm

3 steps

1Step 1: Sort the grades array in ascending order.
2Step 2: Initialize variables for group size and total groups formed.
3Step 3: Iterate through the sorted grades, forming groups of increasing size until the conditions can no longer be met.

solution.py16 lines

1# Full working Python code
2
3def maxGroups(grades):
4    grades.sort()
5    total_groups = 0
6    current_size = 1
7    current_sum = 0
8
9    for grade in grades:
10        current_sum += grade
11        if current_size <= total_groups + 1:
12            total_groups += 1
13            current_size += 1
14            current_sum = 0  # Reset sum for the next group
15
16    return total_groups

ℹ

Complexity note: The sorting step takes O(n log n), and the subsequent iteration through the grades is O(n), making this approach efficient overall.

1Sorting the grades helps in easily forming valid groups.
2The conditions for group formation create a natural sequence of increasing group sizes.

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