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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) {
            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;
    }
}

  • 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
    }
}

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