Question

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

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 $groupSizes$.

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)]

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;
  }
};

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 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;
}

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
  }
}