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This approach involves sorting the elements of the array based on their frequency. We start by counting the frequency of each element. Next, we create a list of these frequencies sorted in decreasing order. By selecting the most frequent elements first, we remove them from the array until at least half of the array size is removed.
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(n) for storing frequencies.
1using System;
2using System.Collections.Generic;
3using System.Linq;
4
5public class Solution {
6 public int MinSetSize(int[] arr) {
7 var frequency = new Dictionary<int, int>();
8 foreach (var num in arr) {
9 if (frequency.ContainsKey(num))
10 frequency[num]++;
11 else
12 frequency[num] = 1;
13 }
14
15 var freqList = frequency.Values.OrderByDescending(x => x).ToList();
16 int removed = 0, setSize = 0, halfSize = arr.Length / 2;
17 foreach (var freq in freqList) {
18 removed += freq;
19 setSize++;
20 if (removed >= halfSize) break;
21 }
22 return setSize;
23 }
24
25 public static void Main(string[] args) {
26 int[] arr = {3,3,3,3,5,5,5,2,2,7};
27 Solution sol = new Solution();
28 Console.WriteLine(sol.MinSetSize(arr));
29 }
30}
31The C# program uses a Dictionary to store the frequency of each element and sorts these in descending order. The algorithm then accumulates the most frequent numbers until at least half of the integer count from the array is removed, specifically tracking the number of integers removed with a variable.
This approach utilizes bucket sort, where frequencies are counted and used as indices in a bucket array to represent how many numbers occur with identical frequency. This helps in efficiently choosing elements to remove, optimizing beyond traditional sorting by directly accessing through indices.
Time Complexity: O(n) because of bucket sorting.
Space Complexity: O(n) due to auxiliary arrays.
1
This C code implements a bucket-based frequency count. It populates a bucket array where the index represents frequency and the value represents how many numbers have that frequency. By accessing these buckets from highest to lowest, we efficiently calculate the number of elements we need to remove from the array to reach at least half its size.