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

SourceWeekly Contest 166 Q2DifficultyMediumRating1267

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

Solution 1: Hash Table or Array

Thinking

People sharing a group size must fill complete groups of that size. \(n \le 500\): bucket indices by \(\textit{groupSize}\), then slice each bucket into contiguous chunks of that length. A solution is guaranteed, so the slices are the groups.

We use a hash table \(g\) to store which people are in each group size \(\textit{groupSize}\). Then we partition each group size into \(k\) equal parts, with each part containing \(\textit{groupSize}\) 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.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\), where \(n\) is the length of \(\textit{groupSizes}\).

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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)]

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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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
    }
}

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