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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.
1from collections import Counter
2
3def minSetSize(arr):
4 frequency = Counter(arr)
5 freq_values = sorted(frequency.values(), reverse=True)
6
7 removed = 0
8 set_size = 0
9 half_size = len(arr) // 2
10 for freq in freq_values:
11 removed += freq
12 set_size += 1
13 if removed >= half_size:
14 break
15 return set_size
16
17arr = [3,3,3,3,5,5,5,2,2,7]
18print(minSetSize(arr))
19The Python solution employs the Counter class from the collections module to map each element to its frequency. These frequencies are sorted in descending order, and the program computes the minimal set size required to remove half of the integers in the array by iteratively subtracting from the sorted frequency list.
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 Java implementation uses a bucket array to count occurrences of each frequency, allowing efficient track of how many numbers have a certain frequency. By decrementally processing these from larger frequencies, we ensure an efficient reduction that minimizes the set size required to remove half the array elements.