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# Group the People Given the Group Size They Belong To

There are `n` people whose **IDs** go from `0` to `n - 1` and each person belongs **exactly** to one group. Given the array `groupSizes` of length `n` telling the group size each person belongs to, return the groups there are and the people's **IDs** each group includes.

You can return any solution in any order and the same applies for IDs. Also, it is guaranteed that there exists at least one solution. 

**Example 1:**

```
Input: groupSizes = [3,3,3,3,3,1,3]
Output: [[5],[0,1,2],[3,4,6]]
Explanation: 
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`

```java
class Solution {
    public List<List<Integer>> groupThePeople(int[] groupSizes) {
        HashMap<Integer, List<Integer>> map = new HashMap<>();
        List<List<Integer>> ans = new ArrayList<>();
        for (int i = 0; i < groupSizes.length; i++) {
            if (!map.containsKey(groupSizes[i])) {
                List<Integer> temp = new ArrayList<>();
                temp.add(i);
                map.put(groupSizes[i], temp);
            } else {
                map.get(groupSizes[i]).add(i);
            }
            if (map.get(groupSizes[i]).size() == groupSizes[i]) {
                ans.add(map.get(groupSizes[i]));
                map.put(groupSizes[i], new ArrayList<Integer>());
            }
        }
        return ans;
    }
}
```


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