human
Solve in Python:
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
assistant
#Here is the Python program you requested.

class Solution:
    def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
        groups = {}
        output = []

        for i in range(len(groupSizes)):
            belongGroupCount = groupSizes[i]
            # print(f'group: {belongGroupCount} | person i: {i}')

            existingGroup = groups.get(belongGroupCount)
            # print(f'existing: {groups}, belong: {belongGroupCount}')
            if existingGroup is None:
                groups[belongGroupCount] = []
                existingGroup = groups.get(belongGroupCount)

            existingGroup.append(i)
            if len(existingGroup) == belongGroupCount:
                output.append(existingGroup)
                groups[belongGroupCount] = []
            else:
                groups[belongGroupCount] = existingGroup

        return output