Group the People Given the Group Size They Belong To

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Roblox

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ArraysGreedy Algorithms

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

Group the People Given the Group Size They Belong To

Medium

Roblox

1 view

Topics:

ArraysGreedy Algorithms

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

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the group sizes in the input array? Are they guaranteed to be positive integers?
Is it guaranteed that for each group size, there will be a number of people that can form complete groups (i.e., the number of people with a given group size is a multiple of that group size)?
If there are multiple possible groupings, is the order of groups within the final output important? Should I return any valid grouping or is there a specific order required?
What should I return if the input array is empty or null?
Are the people represented by integers (indices) from 0 to n-1 or will they be provided as input values alongside the group size?

Brute Force Solution

Approach

The problem involves grouping people based on their group size. The brute force approach explores all possible combinations of people to form groups of the specified sizes, ensuring each group satisfies the size requirement.

Here's how the algorithm would work step-by-step:

Start with an empty list of groups.
Consider the first person and try to include them in a group of their required size by pairing them with other available people.
Check if you can find enough people who require a group of the same size as the first person.
If you can form a valid group, add this group to your list of groups and remove the selected people from the pool of available people.
If you cannot form a valid group for the first person, move on and repeat this process for the next available person.
Continue this process until all people have been considered and placed into valid groups, if possible.
If you can't place everyone into groups, it means there is no valid solution. Return the groups you created if a solution was found.

Code Implementation

def group_the_people(group_sizes):
    people_indices = list(range(len(group_sizes)))
    resulting_groups = []

    while people_indices:
        first_person_index = people_indices[0]
        group_size_required = group_sizes[first_person_index]
        current_group = [first_person_index]
        available_people = people_indices[1:]
        
        # We need to find the others to form a group of the required size
        for person_index in available_people:
            if group_sizes[person_index] == group_size_required:
                current_group.append(person_index)
                if len(current_group) == group_size_required:
                    break
        
        if len(current_group) == group_size_required:
            resulting_groups.append(current_group)

            # Remove the selected people from the available pool.
            people_indices = [person for person in people_indices if person not in current_group]
        else:
            #If we could not make a group of sufficient size, return an empty list
            return []

    return resulting_groups

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each person (n total) to find a suitable group. For each person, it potentially iterates through the remaining people to find others with the same group size requirement to form a complete group. In the worst case, this inner search could involve checking almost all other people, leading to a nested loop-like behavior. Thus, the operations are roughly proportional to n * n/2 which simplifies to O(n²).

Space Complexity

O(N) – The algorithm maintains a list of groups. In the worst-case scenario, each person might belong to a group of size 1, leading to N groups. Consequently, the auxiliary space required to store these groups grows linearly with the number of people, N. This results in a space complexity of O(N) due to the storage of the list of groups each containing individuals.

Optimal Solution

Approach

The best way to solve this is to organize people by their group size. Then, for each group size, we form groups until everyone is assigned.

Here's how the algorithm would work step-by-step:

First, organize everyone based on the size of the group they need to be in.
Then, go through each group size one by one.
For each group size, create groups until you have assigned everyone who belongs to that group size.
Move on to the next group size and repeat the process until everyone is in a group.

Code Implementation

def groupThePeople(group_sizes): 
    # Organize people by the size of the group they belong to.
    groups_by_size = {}
    for person_id, group_size in enumerate(group_sizes): 
        if group_size not in groups_by_size:
            groups_by_size[group_size] = []
        groups_by_size[group_size].append(person_id)

    result_groups = []

    # Iterate through each group size.
    for group_size, person_ids in groups_by_size.items():
        # Create groups of the current size until all people are assigned.
        for i in range(0, len(person_ids), group_size):

            result_groups.append(person_ids[i:i + group_size])

    return result_groups

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n once to group people by their group size, which takes O(n) time. For each group size, it iterates through the list of people belonging to that size, forming groups. The total number of people across all groups sizes is n. Therefore, the group formation process also takes O(n) time. Since both steps are performed sequentially, the overall time complexity is O(n) + O(n), which simplifies to O(n).

Space Complexity

O(N) – The algorithm organizes people based on group size, implying the creation of a data structure (like a hash map or a list of lists) to store people belonging to the same group size. In the worst case, each person might have a unique group size, leading to storage for all N people. Therefore, the auxiliary space required grows linearly with the number of people, N. The additional space needed will hold the N people.

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty list as there are no people to group.
Input array contains group sizes that are not factors of the array's length	The algorithm should gracefully handle cases where a valid grouping isn't possible without producing incorrect or incomplete groups.
Array with only one group size value	Correctly form groups of that single size from the input array's indices.
Array with large input size (e.g., 500 elements), efficiency concerns	Ensure the solution uses a data structure like a hash map to maintain O(n) time complexity for optimal scaling.
All elements in the input array are the same, requiring a single large group.	The algorithm forms a single large group with the indices associated with that group size.
Some group size values are larger than the input array length	Ignore group sizes that are larger than the number of indices available to form a group.
Multiple valid groupings possible; should the algorithm return all of them or just one?	The problem statement should specify whether all valid groupings or a single valid grouping is expected; the provided solution returns just one.
Input array contains zero values for group sizes	Filter out zero values since a group size cannot be zero.