Introduction

In this blog post, we will explore how to solve the LeetCode problem "1282" 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

There are n people that are split into some unknown number of groups. Each person is labeled with a unique ID from 0 to n - 1. You are given an integer array groupSizes, where groupSizes[i] is the size of the group that person i is in. For example, if groupSizes[1] = 3, then person 1 must be in a group of size 3. Return a list of groups such that each person i is in a group of size groupSizes[i]. Each person should appear in exactly one group, and every person must be in a group. If there are multiple answers, return any of them. It is guaranteed that there will be at least one valid solution for the given input. Example 1: Input: groupSizes = [3,3,3,3,3,1,3] Output: [[5],[0,1,2],[3,4,6]] Explanation: The first group is [5]. The size is 1, and groupSizes[5] = 1. The second group is [0,1,2]. The size is 3, and groupSizes[0] = groupSizes[1] = groupSizes[2] = 3. The third group is [3,4,6]. The size is 3, and groupSizes[3] = groupSizes[4] = groupSizes[6] = 3. Other possible solutions are [[2,1,6],[5],[0,4,3]] and [[5],[0,6,2],[4,3,1]]. Example 2: Input: groupSizes = [2,1,3,3,3,2] Output: [[1],[0,5],[2,3,4]] Constraints: groupSizes.length == n 1 <= n <= 500 1 <= groupSizes[i] <= n

Explanation

Here's a breakdown of the solution:

• Group by Size: Create a dictionary (or hash map) where the keys are the group sizes and the values are lists of person IDs that belong to groups of that size.

• Form Groups: Iterate through the dictionary. For each group size, create groups of the specified size using the person IDs associated with it. If the number of people with a particular size is a multiple of the size, distribute evenly; if it is not a direct multiple then create as many groups of the specified size as possible, and the code guarantees a valid solution exists.

• Runtime & Storage Complexity: O(n), where n is the number of people (length of groupSizes). Storage complexity is also O(n).

Code

    def groupThePeople(groupSizes):
    """
    Groups people based on their group sizes.

    Args:
        groupSizes: A list of integers representing the size of the group each person belongs to.

    Returns:
        A list of lists, where each inner list represents a group of people.
    """

    group_map = {}
    for i, size in enumerate(groupSizes):
        if size not in group_map:
            group_map[size] = []
        group_map[size].append(i)

    result = []
    for size, people in group_map.items():
        num_groups = len(people) // size
        for i in range(num_groups):
            result.append(people[i * size:(i + 1) * size])

    return result