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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.
1import java.util.*;
2
3public class ReduceArraySize {
4 public static int minSetSize(int[] arr) {
5 Map<Integer, Integer> frequency = new HashMap<>();
6 for (int num : arr) {
7 frequency.put(num, frequency.getOrDefault(num, 0) + 1);
8 }
9
10 List<Integer> frequencies = new ArrayList<>(frequency.values());
11 frequencies.sort(Collections.reverseOrder());
12
13 int removed = 0, setSize = 0, halfSize = arr.length / 2;
14 for (int freq : frequencies) {
15 removed += freq;
16 setSize++;
17 if (removed >= halfSize) {
18 break;
19 }
20 }
21 return setSize;
22 }
23
24 public static void main(String[] args) {
25 int[] arr = {3,3,3,3,5,5,5,2,2,7};
26 System.out.println(minSetSize(arr));
27 }
28}
29This Java implementation creates a hash map to store the frequency of each element in the array. These frequencies are added to a list, sorted in descending order, and then the program calculates the minimum set needed to remove half of the array's integers by iterating over the sorted frequencies until the required condition is met.
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.
1using System.Collections.Generic;
public class Solution {
public int MinSetSize(int[] arr) {
var frequency = new Dictionary<int, int>();
foreach (var num in arr) {
if (frequency.ContainsKey(num))
frequency[num]++;
else
frequency[num] = 1;
}
int[] bucket = new int[arr.Length + 1];
foreach (var freq in frequency.Values) {
bucket[freq]++;
}
int removed = 0, setSize = 0, halfSize = arr.Length / 2;
for (int i = arr.Length; i > 0; i--) {
while (bucket[i] > 0) {
removed += i;
setSize++;
bucket[i]--;
if (removed >= halfSize) return setSize;
}
}
return setSize;
}
public static void Main(string[] args) {
int[] arr = {3,3,3,3,5,5,5,2,2,7};
Solution sol = new Solution();
Console.WriteLine(sol.MinSetSize(arr));
}
}
The C# code implements a strategy based on bucket sorting. After calculating element frequencies, it populates a bucket array to directly access and subtract frequencies in descending order, enabling a direct and efficient calculation to fulfill the necessary removal condition with minimal set size.