Introduction

In this blog post, we will explore how to solve the LeetCode problem "2358" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the breakdown of the solution:

• Greedy Approach: The core idea is to form groups greedily. We aim to minimize the sum of grades and the number of students in each group to maximize the number of groups we can create. We sort the grades to efficiently pick the smallest grades for earlier groups.
• Iterative Group Formation: We iterate, building groups one by one. In each iteration, we check if adding the next smallest grades to the current group maintains the increasing sum and count conditions.

• Maintaining Conditions: The current_sum and current_count variables track the sum and count of the last formed group. We ensure the next group's sum and count are strictly greater.

• Runtime Complexity: O(n log n), primarily due to sorting the grades. Storage Complexity: O(n), where n is the number of grades because we store the sorted list.

Code

    def max_groups(grades):
    """
    Calculates the maximum number of groups that can be formed from a list of grades,
    satisfying the given conditions.

    Args:
        grades: A list of positive integers representing the grades of students.

    Returns:
        The maximum number of groups that can be formed.
    """
    grades.sort()
    current_sum = 0
    current_count = 0
    groups = 0
    index = 0

    while index < len(grades):
        next_sum = 0
        next_count = 0
        temp_index = index
        while temp_index < len(grades) and next_count <= current_count:
            next_sum += grades[temp_index]
            next_count += 1
            temp_index += 1

        if next_count > current_count and next_sum > current_sum:
            current_sum = next_sum
            current_count = next_count
            groups += 1
            index = temp_index
        else:
            break

    return groups