Sponsored
Sponsored
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.
1function minSetSize(arr) {
2 const frequency = {};
3 arr.forEach(num => {
4 frequency[num] = (frequency[num] || 0) + 1;
5 });
6
7 const freqValues = Object.values(frequency).sort((a, b) => b - a);
8 let removed = 0, setSize = 0, halfSize = arr.length / 2;
9 for (const freq of freqValues) {
10 removed += freq;
11 setSize++;
12 if (removed >= halfSize) {
13 break;
14 }
15 }
16 return setSize;
17}
18
19console.log(minSetSize([3,3,3,3,5,5,5,2,2,7]));
20The JavaScript code defines a function minSetSize which constructs an object to record the frequency of each number in the array. These frequencies are extracted, sorted in descending order, and processed to find the minimal set size ensuring half of the array elements are removed. This is done via iterating the sorted frequency list and reducing the count by their most frequent elements first.
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.