LeetCode 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]]
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]]


  • groupSizes.length == n
  • 1 <= n <= 500
  • 1 <= groupSizes[i] <= n


Group the people(index) by group size. Then create groups for each size of groups, if group size is full, create a new group.

Python Solution

class Solution:
    def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
        results = []
        positions = defaultdict(list)
        for i, group_size in enumerate(groupSizes):
        for key, value in positions.items():
            group = []      
            for i in value:

                if len(group) == key:
                    group = []
        return results
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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