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`

```
struct Solution;
type Person = (i32, usize);
impl Solution {
fn group_the_people(group_sizes: Vec<i32>) -> Vec<Vec<i32>> {
let mut people: Vec<Person> = vec![];
let mut res: Vec<Vec<i32>> = vec![];
for (id, &group_size) in group_sizes.iter().enumerate() {
people.push((group_size, id))
}
people.sort_unstable();
let mut group: Vec<i32> = vec![];
for p in people {
group.push(p.1 as i32);
if group.len() == p.0 as usize {
res.push(group);
group = vec![];
}
}
res
}
}
#[test]
fn test() {
let group_sizes = vec![3, 3, 3, 3, 3, 1, 3];
let res = vec_vec_i32![[5], [0, 1, 2], [3, 4, 6]];
assert_eq!(Solution::group_the_people(group_sizes), res);
let group_sizes = vec![2, 1, 3, 3, 3, 2];
let res = vec_vec_i32![[1], [0, 5], [2, 3, 4]];
assert_eq!(Solution::group_the_people(group_sizes), res);
}
```