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

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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

Python3

class Solution:
    def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
        mp = defaultdict(list)
        for i, x in enumerate(groupSizes):
            mp[x].append(i)
        res = []
        for x, indexes in mp.items():
            l = len(indexes)
            for i in range(0, l, x):
                res.append(indexes[i: i + x])
        return res

Java

class Solution {
    public List<List<Integer>> groupThePeople(int[] groupSizes) {
        Map<Integer, List<Integer>> mp = new HashMap<>();
        for (int i = 0; i < groupSizes.length; ++i) {
            mp.computeIfAbsent(groupSizes[i], k -> new ArrayList<>()).add(i);
        }
        List<List<Integer>> res = new ArrayList<>();
        for (Map.Entry<Integer, List<Integer>> entry : mp.entrySet()) {
            int x = entry.getKey();
            List<Integer> indexes = entry.getValue();
            for (int i = 0; i < indexes.size(); i += x) {
                res.add(new ArrayList<>(indexes.subList(i, i + x)));
            }
        }
        return res;
    }
}

C++

class Solution {
public:
    vector<vector<int>> groupThePeople(vector<int>& groupSizes) {
        unordered_map<int, vector<int>> mp;
        for (int i = 0; i < groupSizes.size(); ++i) mp[groupSizes[i]].push_back(i);
        vector<vector<int>> res;
        for (auto& entry : mp)
        {
            int x = entry.first;
            auto indexes = entry.second;
            for (int i = 0; i < indexes.size(); i += x)
            {
                vector<int> t(indexes.begin() + i, indexes.begin() + i + x);
                res.push_back(t);
            }
        }
        return res;
    }
};

Go

func groupThePeople(groupSizes []int) [][]int {
	mp := make(map[int][]int)
	for i, x := range groupSizes {
		mp[x] = append(mp[x], i)
	}
	var res [][]int
	for x, indexes := range mp {
		for i := 0; i < len(indexes); i += x {
			res = append(res, indexes[i:i+x])
		}
	}
	return res
}

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