# LeetCode 1282. Group the People Given the Group Size They Belong To

## Description

https://leetcode.com/problems/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 = 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: [,[0,1,2],[3,4,6]]
Explanation:
The first group is . The size is 1, and groupSizes = 1.
The second group is [0,1,2]. The size is 3, and groupSizes = groupSizes = groupSizes = 3.
The third group is [3,4,6]. The size is 3, and groupSizes = groupSizes = groupSizes = 3.
Other possible solutions are [[2,1,6],,[0,4,3]] and [,[0,6,2],[4,3,1]].
```

Example 2:

```Input: groupSizes = [2,1,3,3,3,2]
Output: [,[0,5],[2,3,4]]
```

Constraints:

• `groupSizes.length == n`
• `1 <= n <= 500`
• `1 <= groupSizes[i] <= n`

## Explanation

Group the people(index) by group size. Then create groups for each size of groups, if group size is full, create a new group.

## Python Solution

``````class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
results = []

positions = defaultdict(list)

for i, group_size in enumerate(groupSizes):
positions[group_size].append(i)

for key, value in positions.items():
group = []

for i in value:
group.append(i)

if len(group) == key:
results.append(group)
group = []

return results``````
• Time Complexity: O(N).
• Space Complexity: O(N).