# Leetcode 1282. Group the People Given the Group Size They Belong To

[medium]

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`

[Think]

- create a map, and put same group size people into the list
- if list size match the group size number, add it into the result list, and remove the list from map
- In this answer, it had at least one answer, so no need to worry if anyone didn’t join the group
- We could know that after the end of loop, the map should be empty

[Java]