We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsCompaniesDisclaimersPrivacyTermsContactAbout Us

Maximum Number of Groups Entering a Competition

Number: 2437

Difficulty: Medium

Paid? No

Companies: N/A

Problem Description

Given an array of positive integers representing student grades, partition all students into ordered non-empty groups such that:

  • The sum of grades in the iᵗʰ group is less than that of the (i + 1)ᵗʰ group.
  • The number of students in the iᵗʰ group is less than that of the (i + 1)ᵗʰ group.

Return the maximum number of groups that can be formed.

Key Insights

  • Since all grades are positive, any increase in group size guarantees the possibility of an increasing sum if grouped judiciously.
  • To maximize the number of groups, assign the smallest possible group size to the first group (which is 1), the next smallest (2) to the second, and so on.
  • The total number of students required for k groups is 1 + 2 + ... + k = k * (k + 1) / 2.
  • The maximum k is the largest integer such that k*(k+1)/2 is less than or equal to the total number of students (n = grades.length).

Space and Time Complexity

Time Complexity: O(1) – The computation uses a mathematical formula derived from n. Space Complexity: O(1) – Only constant extra space is used.

Solution

The key observation is that forming groups with sizes 1, 2, 3, ... will automatically satisfy the condition that each group's size is less than the next. Given that all numbers are positive, the smallest possible sum increases as the group size increases. Therefore, maximize the number of groups by finding the maximum integer k such that the sum of the first k natural numbers (k * (k + 1) / 2) does not exceed n (the length of the grades array).

The solution involves solving k*(k+1)/2 ≤ n. This can be done by recognizing it is a quadratic inequality in k and solving it with the quadratic formula. The answer is given by: k = floor((sqrt(1+8*n) - 1) / 2).

This approach guarantees that both conditions of the problem can be satisfied by careful assignment of the students to groups.

Code Solutions

import math

def maximumGroups(grades):
    # Total number of students
    n = len(grades)
    # Solve the inequality k*(k+1)/2 <= n for k using the quadratic formula
    # k = (sqrt(1 + 8*n) - 1) / 2, we take the floor of the result
    k = int((math.sqrt(1 + 8 * n) - 1) // 2)
    return k

# Example usage:
grades_example1 = [10,6,12,7,3,5]
print(maximumGroups(grades_example1))  # Output: 3
← Back to All Questions