## DEV Community

Abhishek Chaudhary

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# 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`

SOLUTION:

``````class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
numpeople = {}
for i, gsize in enumerate(groupSizes):
numpeople[gsize] = numpeople.get(gsize, []) + [i]
op = []
for size, people in numpeople.items():
n = len(people)
for i in range(0, n, size):
op.append(people[i:i+size])
return op
``````