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

Description

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

Solutions

• class Solution {
public List<List<Integer>> groupThePeople(int[] groupSizes) {
int n = groupSizes.length;
List<Integer>[] g = new List[n + 1];
Arrays.setAll(g, k -> new ArrayList<>());
for (int i = 0; i < n; ++i) {
}
List<List<Integer>> ans = new ArrayList<>();
for (int i = 0; i < g.length; ++i) {
List<Integer> v = g[i];
for (int j = 0; j < v.size(); j += i) {
}
}
return ans;
}
}

• class Solution {
public:
vector<vector<int>> groupThePeople(vector<int>& groupSizes) {
int n = groupSizes.size();
vector<vector<int>> g(n + 1);
for (int i = 0; i < n; ++i) g[groupSizes[i]].push_back(i);
vector<vector<int>> ans;
for (int i = 0; i < g.size(); ++i) {
for (int j = 0; j < g[i].size(); j += i) {
vector<int> t(g[i].begin() + j, g[i].begin() + j + i);
ans.push_back(t);
}
}
return ans;
}
};

• class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
g = defaultdict(list)
for i, v in enumerate(groupSizes):
g[v].append(i)
return [v[j : j + i] for i, v in g.items() for j in range(0, len(v), i)]


• func groupThePeople(groupSizes []int) [][]int {
n := len(groupSizes)
g := make([][]int, n+1)
for i, v := range groupSizes {
g[v] = append(g[v], i)
}
ans := [][]int{}
for i, v := range g {
for j := 0; j < len(v); j += i {
ans = append(ans, v[j:j+i])
}
}
return ans
}

• function groupThePeople(groupSizes: number[]): number[][] {
const res = [];
const map = new Map<number, number[]>();
const n = groupSizes.length;
for (let i = 0; i < n; i++) {
const size = groupSizes[i];
map.set(size, [...(map.get(size) ?? []), i]);
const arr = map.get(size);
if (arr.length === size) {
res.push(arr);
map.set(size, []);
}
}
return res;
}


• use std::collections::HashMap;
impl Solution {
pub fn group_the_people(group_sizes: Vec<i32>) -> Vec<Vec<i32>> {
let mut res = vec![];
let mut map = HashMap::new();
for i in 0..group_sizes.len() {
let size = group_sizes[i] as usize;
let arr = map.entry(size).or_insert(vec![]);
arr.push(i as i32);
if arr.len() == size {
res.push(arr.clone());
arr.clear();
}
}
res
}
}