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.
1function topKFrequent(nums, k) {
2 const freqMap = {};
3 for (const num of nums) {
4 freqMap[num] = (freqMap[num] || 0) + 1;
5 }
6
7 const minHeap = Object.keys(freqMap).sort((a, b) => freqMap[b] - freqMap[a]).slice(0, k);
8
9 return minHeap.map(Number);
10}
11
12const nums = [1, 1, 1, 2, 2, 3];
13const k = 2;
14console.log(topKFrequent(nums, k));
15We use an object to count frequencies, sort keys by frequency, and take the top k.
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).
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 List<Integer>[] buckets = new List[nums.length + 1];
11 for (int key : freqMap.keySet()) {
12 int frequency = freqMap.get(key);
13 if (buckets[frequency] == null) {
14 buckets[frequency] = new ArrayList<>();
15 }
16 buckets[frequency].add(key);
17 }
18
19 List<Integer> resultList = new ArrayList<>();
20 for (int pos = buckets.length - 1; pos >= 0 && resultList.size() < k; pos--) {
21 if (buckets[pos] != null) {
22 resultList.addAll(buckets[pos]);
23 }
24 }
25
26 int[] result = new int[k];
27 for (int i = 0; i < k; i++) {
28 result[i] = resultList.get(i);
29 }
30 return result;
31 }
32
33 public static void main(String[] args) {
34 Solution sol = new Solution();
35 int[] nums = {1, 1, 1, 2, 2, 3};
36 int k = 2;
37 int[] result = sol.topKFrequent(nums, k);
38 System.out.println(Arrays.toString(result));
39 }
40}
41By clustering elements into frequency buckets, the most frequented elements appear in the last non-empty buckets.