Group the People Given the Group Size They Belong To

Problem

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 (Java)

class Solution {
    public List<List<Integer>> groupThePeople(int[] groupSizes) {
        Map<Integer, List<Integer>> map = new HashMap<>();
        for (int i = 0; i < groupSizes.length; i++) {
            List<Integer> list = map.getOrDefault(groupSizes[i], new ArrayList<>());
            list.add(i);
            map.put(groupSizes[i], list);
        }
        List<List<Integer>> result = new ArrayList<>();
        for (Map.Entry<Integer, List<Integer>> entry : map.entrySet()) {
            List<Integer> list = entry.getValue();
            int i = 0;
            do {
                result.add(list.subList(i, i + entry.getKey()));
                i += entry.getKey();
            } while (i + entry.getKey() <= list.size());
        }
        return result;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).