# Leetcode - Group the People Given the Group Size They Belong To Solution

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 in Python

```
from collections import defaultdict
class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
groups = defaultdict(list)
for x,y in enumerate(groupSizes):
groups[y].append(x)
output = []
for size,arr in groups.items():
for i in range(0,len(arr),size):
output.append(arr[i:i+size])
return output
```