This approach uses a hash map to count the frequency of each element. We then use a min-heap to keep track of the top k elements.
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(n) for storing frequencies.
1import java.util.*;
2
3class Solution {
4 public int[] topKFrequent(int[] nums, int k) {
5 Map<Integer, Integer> freqMap = new HashMap<>();
6 for (int num : nums) {
7 freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);
8 }
9
10 PriorityQueue<Integer> minHeap = new PriorityQueue<>((a, b) -> freqMap.get(a) - freqMap.get(b));
11 for (int num : freqMap.keySet()) {
12 minHeap.add(num);
13 if (minHeap.size() > k) {
14 minHeap.poll();
15 }
16 }
17
18 int[] result = new int[k];
19 for (int i = k - 1; i >= 0; i--) {
20 result[i] = minHeap.poll();
21 }
22 return result;
23 }
24
25 public static void main(String[] args) {
26 Solution sol = new Solution();
27 int[] nums = {1,1,1,2,2,3};
28 int k = 2;
29 int[] result = sol.topKFrequent(nums, k);
30 System.out.println(Arrays.toString(result));
31 }
32}
33HashMap is used to record the frequency of each element, and a min heap of size k keeps track of the top k elements based on frequency.
This approach involves using bucket sort where we create buckets for frequency counts and then extract the top k frequent elements.
Time Complexity: O(n + k).
Space Complexity: O(n).
1from collections import Counter
2
3def topKFrequent(nums, k):
4 count = Counter(nums)
5 buckets = [[] for _ in range(len(nums) + 1)]
6
7 for num, freq in count.items():
8 buckets[freq].append(num)
9
10 result = []
11 for i in range(len(buckets) - 1, 0, -1):
12 for num in buckets[i]:
13 result.append(num)
14 if len(result) == k:
15 return result
16
17if __name__ == '__main__':
18 nums = [1, 1, 1, 2, 2, 3]
19 k = 2
20 print(topKFrequent(nums, k))
21Even with Python, we leverage frequency count with buckets to directly access and collect top k frequent elements.