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Group the People Given the Group Size They Belong To

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 Explanation for Group the People Given the Group Size They Belong To

This problem requires grouping people based on the size of the group they belong to. The input is an array groupSizes where groupSizes[i] represents the desired group size for person i. The output should be a list of groups, each containing the correct number of people.

The most efficient approach leverages a hash table (or dictionary in Python) to group people by their desired group size. We iterate through the groupSizes array. For each person, we append their index (ID) to the list associated with their group size in the hash table.

After processing all people, we iterate through the hash table. For each group size, we partition the list of people into sub-lists, each of the specified size. These sub-lists represent the final groups.

Algorithm:

  1. Initialization: Create a hash table (or dictionary) to store lists of people for each group size. The keys will be the group sizes, and the values will be lists of person IDs.

  2. Grouping: Iterate through the groupSizes array. For each person i, add i to the list associated with groupSizes[i] in the hash table.

  3. Partitioning: Iterate through the hash table. For each group size size and its corresponding list people, partition the list into sub-lists of size size. This involves creating new lists, each containing size consecutive elements from people. Add these sub-lists to the final result.

  4. Return: Return the list of groups.

Time Complexity: O(N), where N is the number of people. We iterate through the groupSizes array once (O(N)) and then iterate through the hash table, where the total number of elements across all lists in the hash table is also O(N). The partitioning step takes linear time proportional to the total number of people.

Space Complexity: O(N), primarily due to the hash table which can store up to N entries in the worst-case scenario (all people in separate groups).

Code Examples:

The provided code examples demonstrate the solution in various programming languages, each following the algorithm described above. The key differences lie in the specific syntax and data structures used (e.g., defaultdict in Python, HashMap in Java, etc.), but the underlying logic remains consistent across all implementations. The use of list slicing or sublist methods (subList in Java, slicing in Python) is crucial for efficient partitioning.

The choice between using a hash table (or dictionary) and a simple array depends on the programming language and its capabilities. If the range of possible group sizes is known and relatively small, an array might be more efficient due to potentially faster lookups. However, a hash table provides more flexibility and generally handles diverse group sizes more effectively.