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 == n1 <= n <= 5001 <= groupSizes[i] <= n
Solutions
Solution 1: Hash Table or Array
We use a hash table g to store which people are in each group size groupSize. Then we partition each group size into k equal parts, with each part containing groupSize people.
Since the range of n in the problem is small, we can also directly create an array of size n+1 to store the data, which is more efficient.
Time complexity is O(n), and space complexity is O(n). Here, n is the length of groupSizes.
PythonJavaC++GoTypeScriptRust
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)](code-box)
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) {
g[groupSizes[i]].add(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) {
ans.add(v.subList(j, j + i));
}
}
return ans;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
function groupThePeople(groupSizes: number[]): number[][] {
const n: number = groupSizes.length;
const g: number[][] = Array.from({ length: n + 1 }, () => []);
for (let i = 0; i < groupSizes.length; i++) {
const size: number = groupSizes[i];
g[size].push(i);
}
const ans: number[][] = [];
for (let i = 1; i <= n; i++) {
const group: number[] = [];
for (let j = 0; j < g[i].length; j += i) {
group.push(...g[i].slice(j, j + i));
ans.push([...group]);
group.length = 0;
}
}
return ans;
}(code-box)
impl Solution {
pub fn group_the_people(group_sizes: Vec<i32>) -> Vec<Vec<i32>> {
let n: usize = group_sizes.len();
let mut g: Vec<Vec<usize>> = vec![Vec::new(); n + 1];
for (i, &size) in group_sizes.iter().enumerate() {
g[size as usize].push(i);
}
let mut ans: Vec<Vec<i32>> = Vec::new();
for (i, v) in g.into_iter().enumerate() {
for j in (0..v.len()).step_by(i.max(1)) {
ans.push(
v[j..(j + i).min(v.len())]
.iter()
.map(|&x| x as i32)
.collect(),
);
}
}
ans
}
}(code-box)
