Group the People Given the Group Size They Belong To

# 1282. 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
```

/**
 * @param {number[]} groupSizes
 * @return {number[][]}
 */
 var groupThePeople = function(groupSizes) {
    const out = [];
    const map = new Map();
      
    for (let i = 0; i < groupSizes.length; i++) {
      const groupSize = groupSizes[i];
      
      const groupList = map.get(groupSize);
      (groupList) ? groupList.push(i) : map.set(groupSize, [i]);
      
      if (map.get(groupSize).length === groupSize) {
        out.push(map.get(groupSize));
        map.delete(groupSize);
      }
    }
      
    return out;  
  };