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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.
1#include <stdio.h>
2#include <stdlib.h>
3
4int compare(const void *a, const void *b) {
5 return *(int*)b - *(int*)a;
6}
7
8int minSetSize(int* arr, int arrSize) {
9 int frequency[100001] = {0};
10 for (int i = 0; i < arrSize; i++) {
11 frequency[arr[i]]++;
12 }
13
14 int freqCount[100001] = {0};
15 int freqSize = 0;
16 for (int i = 0; i < 100001; i++) {
17 if (frequency[i] > 0) {
18 freqCount[freqSize++] = frequency[i];
19 }
20 }
21
22 qsort(freqCount, freqSize, sizeof(int), compare);
23
24 int sum = 0, setSize = 0;
25 for (int i = 0; i < freqSize; i++) {
26 sum += freqCount[i];
27 setSize++;
28 if (sum >= arrSize / 2) {
29 return setSize;
30 }
31 }
32 return setSize;
33}
34
35int main() {
36 int arr[] = {3,3,3,3,5,5,5,2,2,7};
37 int arrSize = sizeof(arr)/sizeof(arr[0]);
38 printf("%d\n", minSetSize(arr, arrSize));
39 return 0;
40}
41The code defines a function minSetSize that calculates the minimum set size required to remove at least half of the integers in the array. It uses an array to count the frequency of each element, sorts this frequency array, and removes elements starting with the most frequent to fulfill the condition. The function returns the size of the set needed.
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
In JavaScript, the approach uses bucket sorting to efficiently track the number of elements with the same frequency. This index-based method allows efficiently traversing from highest to smallest frequency, ensuring that we quickly achieve the requisite element removal while minimizing the necessary set size.