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).
1#include <iostream>
2#include <vector>
3#include <unordered_map>
4#include <algorithm>
5
6using namespace std;
7
8vector<int> topKFrequent(vector<int>& nums, int k) {
9 unordered_map<int, int> freqMap;
10 vector<vector<int>> buckets(nums.size() + 1);
11
12 for (int num : nums) {
13 freqMap[num]++;
14 }
15
16 for (auto& p : freqMap) {
17 buckets[p.second].push_back(p.first);
18 }
19
20 vector<int> result;
21 for (int i = buckets.size() - 1; i >= 0 && result.size() < k; --i) {
22 for (int num : buckets[i]) {
23 result.push_back(num);
24 if (result.size() == k) break;
25 }
26 }
27 return result;
28}
29
30int main() {
31 vector<int> nums = {1, 1, 1, 2, 2, 3};
32 int k = 2;
33 vector<int> result = topKFrequent(nums, k);
34 for (int num : result) {
35 cout << num << " ";
36 }
37 return 0;
38}
39Frequency occurrences are placed in buckets. The largest buckets represent the most frequent elements.