1282

1282. Group the People Given the Group Size They Belong To (Medium)

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

Companies: C3 IoT, Apple, Roblox

Related Topics:
Array, Hash Table

Similar Questions:

• Maximum Number of Groups With Increasing Length (Hard)

Solution 1.

// OJ: https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<vector<int>> groupThePeople(vector<int>& A) {
        unordered_map<int, vector<int>> m; // a map from group size to the group members
        vector<vector<int>> ans;
        for (int i = 0; i < A.size(); ++i) {
            m[A[i]].push_back(i);
            if (m[A[i]].size() == A[i]) { // if this group is full, add it into answer
                ans.push_back(m[A[i]]);
                m[A[i]].clear();
            }
        }
        return ans;
    }
};

Solution 2.

The worst case for space complexity analysis is where we have the input in the form of [1, 2, 2, 3, 3, 3, ...]. Assume there are k distinct sizes, then k * (k + 1) / 2 = N, so the space complexity is O(sqrt(N)).

// OJ: https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(sqrt(N))
class Solution {
public:
    vector<vector<int>> groupThePeople(vector<int>& A) {
        unordered_map<int, int> m; // a map from group size to the corresponding index in answer
        vector<vector<int>> ans;
        for (int i = 0; i < A.size(); ++i) {
            int sz = A[i];
            if (m.count(sz) == 0 || ans[m[sz]].size() == sz) {
                m[sz] = ans.size();
                ans.emplace_back();
            }
            ans[m[sz]].push_back(i);
        }
        return ans;
    }
};