## 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[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

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).