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Maximum Number of Groups Entering a Competition

You are given a positive integer array grades which represents the grades of students in a university. You would like to enter all these students into a competition in ordered non-empty groups, such that the ordering meets the following conditions:

  • The sum of the grades of students in the ith group is less than the sum of the grades of students in the (i + 1)th group, for all groups (except the last).
  • The total number of students in the ith group is less than the total number of students in the (i + 1)th group, for all groups (except the last).

Return the maximum number of groups that can be formed.

 

Example 1:

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

The problem asks to find the maximum number of groups that can be formed from a given array of student grades, subject to two constraints: the sum of grades and the number of students in each group must strictly increase with the group index.

The core idea is to realize that the total number of students (n) and the maximum number of groups (k) are related by an inequality derived from the constraints. Let's analyze this relationship.

If we have k groups, the number of students in each group forms an arithmetic progression: 1, 2, 3, ..., k. The total number of students is the sum of this arithmetic series:

1 + 2 + 3 + ... + k = k * (k + 1) / 2

This represents the minimum number of students needed to form k groups. However, since the sum of grades also needs to increase, the total number of students required might be even higher. Consider the worst-case scenario where we have k groups, and the group sizes are at least 1, 2, 3,... k. The minimum sum of grades for the kth group must be strictly greater than the sum of grades in the (k-1)th group. We can say that the lower bound of the total grades is proportional to the square of k. Thus the total number of students, n, must be at least k * (k + 1) / 2.

Let's use x to represent the number of groups. The total number of students is at least x(x+1)/2. However, this is a lower bound, and because the sum of grades must also increase, it's likely the actual student count is higher. In the worst-case scenario, the sum of grades could be greater than the lower bound. The condition we aim for is x(x + 1) / 2 <= n.

We can solve this inequality using binary search to find the maximum possible integer value of x (number of groups).

Time and Space Complexity Analysis

  • Time Complexity: O(log n), dominated by the binary search. Binary search takes logarithmic time with respect to the input size (n).
  • Space Complexity: O(1). The solution uses constant extra space regardless of the input size.

Code Explanation (Python Example)

The Python solution utilizes the bisect_right function from the bisect module, which performs a binary search. It directly finds the index where the condition x * (x + 1) / 2 <= n (or equivalently, x*(x+1) <= 2n) is violated. Subtracting 1 from the result gives the maximum number of groups. The lambda function within bisect_right efficiently calculates the x*(x+1) value for each x within the search space.

Other language solutions use a manual binary search loop achieving the same outcome. They initialize l and r and iteratively refine the range until the maximum value is found. The core logic remains consistent across languages.