K most frequent numbers in array - C# practice - using LINQ - workable solution - underneath LINQ OrderByDescending time complexity - ?

KMostFrequentNumbersInArray1.cs

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace KMostFrequentNumberInArray
{
    /*
     * Find k most frequent numbers in the array
       Problem:
       // nums = [5, 3, 1, 1, 1, 3, 73, 1]
       // k = 1
       // return [1]
 
       // k = 2
       // return [1, 3]
 
       // k = 3

       // return [1, 3, 5]


       First, go through the array once, and keep the count for every distinct value:

       We can use LINK to call order by and then get Top k values.

     */
    class Program
    {
        static void Main(string[] args)
        {
            int[] nums = new int[] { 5, 3, 1, 1, 1, 3, 73, 1 };

            IList<int> test1 = kMostFrequentNumberInArray(nums, 1);
            IList<int> test2 = kMostFrequentNumberInArray(nums, 2);
            IList<int> test3 = kMostFrequentNumberInArray(nums, 3); 
        }

        /*
         * K most frequent numbers in the array 
         * Go through the array once to record the count of each distinct number; 
         * And then, use LINQ calls to get top k most frequent number
         */
        public static IList<int> kMostFrequentNumberInArray(int[] arr, int k)
        {
            IList<int> res = new List<int>();
            if (arr == null || arr.Length == 0)
                return res;

            Dictionary<int, int> numberCounts = new Dictionary<int, int>(); 

            for (int i = 0; i < arr.Length; i++)
            {
                int tmp = arr[i];
                if (numberCounts.ContainsKey(tmp))
                {
                    numberCounts[tmp] += 1; 
                }
                else
                {
                    numberCounts.Add(tmp, 1); 
                }
            }

            var selected = numberCounts.OrderByDescending(x => x.Value).Take(k); 
            foreach(var item in selected)
            {
                res.Add(item.Key); 
            }

            return res; 
        }
    }
}