#2358

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Maximum Number of Groups Entering a Competition

Determine the maximum number of ordered non-empty student groups for a competition using grades array analysis and binary search.

array math binary search greedy

Problem Statement

You are given an array of positive integers grades representing student scores. You need to partition the students into sequential non-empty groups such that each group's total grades exceed the previous group's total, and each subsequent group contains more students than the prior one.

Return the maximum number of such groups that can be formed. For example, given grades = [10,6,12,7,3,5], you can create 3 valid groups: [12], [6,7], [10,3,5]. The goal is to find the highest number of groups satisfying these constraints.

Examples

Example 1

Input: grades = [10,6,12,7,3,5]

Output: 3

The following is a possible way to form 3 groups of students:

1st group has the students with grades = [12]. Sum of grades: 12. Student count: 1
2nd group has the students with grades = [6,7]. Sum of grades: 6 + 7 = 13. Student count: 2
3rd group has the students with grades = [10,3,5]. Sum of grades: 10 + 3 + 5 = 18. Student count: 3 It can be shown that it is not possible to form more than 3 groups.

Example 2

Input: grades = [8,8]

Output: 1

We can only form 1 group, since forming 2 groups would lead to an equal number of students in both groups.

Constraints

1 <= grades.length <= 105
1 <= grades[i] <= 105

Solution Approach

Sort and Prepare

Sort the grades array in ascending order to facilitate forming groups where the sum of grades in each group can grow naturally. This step reduces the complexity of checking possible groupings and ensures that smaller grades can be combined efficiently.

Binary Search over Group Count

Apply binary search on the number of groups, from 1 to the theoretical maximum based on the total number of students. For each mid value, check if it is possible to assign students into that many groups while respecting increasing size constraints and sum conditions.

Validate Candidate Groups

To validate a candidate number of groups, incrementally assign the smallest remaining grades to the current group, ensuring each group has more students than the previous one. If all students can be assigned without violating the size and sum conditions, the candidate is valid; otherwise, reduce the number of groups.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n log n + log n * n) due to sorting and binary search checks over group counts. Space complexity is O(1) beyond input storage, as the validation uses constant extra space.

What Interviewers Usually Probe

Sorting grades can simplify grouping checks.
Binary search may hint at optimizing the number of groups.
Tracking cumulative group sizes ensures valid non-decreasing sequences.

Common Pitfalls or Variants

Common pitfalls

Failing to sort grades first increases complexity and may miss valid groupings.
Not enforcing that each group has more students than the previous group leads to incorrect counts.
Checking only sums without tracking group sizes can produce invalid configurations.

Follow-up variants

Limit group sum differences to a maximum value.
Allow only contiguous student sequences for groups.
Change constraints so each group must have strictly increasing sum by at least a fixed delta.

FAQ

What is the key pattern for Maximum Number of Groups Entering a Competition?

The problem primarily uses binary search over the valid answer space combined with greedy assignment of students.

Can we form groups without sorting grades?

Sorting helps guarantee that smaller grades can fill initial groups, simplifying the validation of non-decreasing sums and group sizes.

How do we know a group count is valid?

A candidate group count is valid if students can be assigned incrementally so each group has more students than the previous and higher total grades.

What is the time complexity of this solution?

Sorting takes O(n log n) and binary search with validation takes O(log n * n), resulting in overall O(n log n) efficiency.

Does the binary search guarantee the maximum number of groups?

Yes, binary search narrows down to the largest number of groups where the assignment rules hold, ensuring an optimal solution.

terminal

Solution

Solution 1: Greedy + Binary Search

Observing the conditions in the problem, the number of students in the $i$-th group must be less than that in the $(i+1)$-th group, and the total score of students in the $i$-th group must be less than that in the $(i+1)$-th group. We only need to sort the students by their scores in ascending order, and then assign $1$, $2$, ..., $k$ students to each group in order. If the last group does not have enough students for $k$, we can distribute these students to the previous last group.

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class Solution:
    def maximumGroups(self, grades: List[int]) -> int:
        n = len(grades)
        return bisect_right(range(n + 1), n * 2, key=lambda x: x * x + x) - 1

Continue Practicing

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