## C# || How To Group People Given The Group Size They Belong To Using C#

The following is a module with functions which demonstrates how to group people given the group size they belong to using C#.

1. Group The People – Problem Statement

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]] ```

2. Group The People – Solution

The following is a solution which demonstrates how to group people given the group size they belong to.

``` 2. Group The People - Solution C# // ============================================================================ // Author: Kenneth Perkins // Date: Feb 14, 2024 // Taken From: http://programmingnotes.org/ // File: Solution.cs // Description: Demonstrates how to group people given group they belong to // ============================================================================ public class Solution { public IList<IList<int>> GroupThePeople(int[] groupSizes) { var ans = new List<IList<int>>(); // A map from group size to the list of indices that are there in the group. var szToGroup = new Dictionary<int, List<int>>(); for (int i = 0; i < groupSizes.Length; i++) { if (!szToGroup.ContainsKey(groupSizes[i])) { szToGroup.Add(groupSizes[i], new List<int>()); } var group = szToGroup[groupSizes[i]]; group.Add(i); // When the list size equals the group size, empty it and store it in the answer. if (group.Count == groupSizes[i]) { ans.Add(group); szToGroup.Remove(groupSizes[i]); } } return ans; } }// http://programmingnotes.org/ 12345678910111213141516171819202122232425262728293031 // ============================================================================//    Author: Kenneth Perkins//    Date:   Feb 14, 2024//    Taken From: http://programmingnotes.org///    File:  Solution.cs//    Description: Demonstrates how to group people given group they belong to// ============================================================================public class Solution {    public IList<IList<int>> GroupThePeople(int[] groupSizes) {        var ans = new List<IList<int>>();         // A map from group size to the list of indices that are there in the group.        var szToGroup = new Dictionary<int, List<int>>();        for (int i = 0; i < groupSizes.Length; i++) {            if (!szToGroup.ContainsKey(groupSizes[i])) {                szToGroup.Add(groupSizes[i], new List<int>());            }             var group = szToGroup[groupSizes[i]];            group.Add(i);             // When the list size equals the group size, empty it and store it in the answer.            if (group.Count == groupSizes[i]) {                ans.Add(group);                szToGroup.Remove(groupSizes[i]);            }        }         return ans;    }}// http://programmingnotes.org/ ```

QUICK NOTES:
The highlighted lines are sections of interest to look out for.

The code is heavily commented, so no further insight is necessary. If you have any questions, feel free to leave a comment below.

Once compiled, you should get this as your output for the example cases:

``` [[0,1,2],[5],[3,4,6]] [[1],[2,3,4],[0,5]] ```