# Algo. Group the People Given the Group Size They Belong To

Source

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

``````public class Solution {
public IList<IList<int>> GroupThePeople(int[] groupSizes) {
var res = new List<IList<int>>();
var dict = new Dictionary<int, List<int>>();
for (int i = 0; i < groupSizes.Length; i++) {
if (dict.Keys.Contains(groupSizes[i])){
var length = dict[groupSizes[i]].Count;
if (length < groupSizes[i]) {
}
}
else
{