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

Source

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

Solution

public class Solution {
    public IList<IList<int>> GroupThePeople(int[] groupSizes) {
      var res = new List<IList<int>>();
      var dict = new Dictionary<int, List<int>>();
      for (int i = 0; i < groupSizes.Length; i++) {
          if (dict.Keys.Contains(groupSizes[i])){
              var length = dict[groupSizes[i]].Count;
              if (length < groupSizes[i]) {
                  dict[groupSizes[i]].Add(i);
              }              
          }
          else
          {
              dict.Add(groupSizes[i], new List<int>());
              dict[groupSizes[i]].Add(i);
          }
          
          if (groupSizes[i] == dict[groupSizes[i]].Count) {
              res.Add(dict[groupSizes[i]]);
              dict.Remove(groupSizes[i]);
          } 
      }
        
      return res;
    }
}