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.
1#include <iostream>
2#include <vector>
3#include <unordered_map>
4#include <queue>
5
6using namespace std;
7
8vector<int> topKFrequent(vector<int>& nums, int k) {
9 unordered_map<int, int> freqMap;
10 for (int num : nums) {
11 freqMap[num]++;
12 }
13
14 auto compare = [&](int a, int b) { return freqMap[a] > freqMap[b]; };
15 priority_queue<int, vector<int>, decltype(compare)> minHeap(compare);
16
17 for (auto& p : freqMap) {
18 minHeap.push(p.first);
19 if (minHeap.size() > k) {
20 minHeap.pop();
21 }
22 }
23
24 vector<int> result;
25 while (!minHeap.empty()) {
26 result.push_back(minHeap.top());
27 minHeap.pop();
28 }
29 return result;
30}
31
32int main() {
33 vector<int> nums = {1, 1, 1, 2, 2, 3};
34 int k = 2;
35 vector<int> result = topKFrequent(nums, k);
36 for (int num : result) {
37 cout << num << " ";
38 }
39 return 0;
40}
41We use a hash map to count frequencies, then use a min-heap of size k to keep track of the top k elements.
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).
1function topKFrequent(nums, k) {
2 const freqMap = {};
3 nums.forEach(num => {
4 freqMap[num] = (freqMap[num] || 0) + 1;
5 });
6
7 const buckets = Array(nums.length + 1).fill().map(() => []);
8 Object.keys(freqMap).forEach(num => {
9 buckets[freqMap[num]].push(parseInt(num));
10 });
11
12 const result = [];
13 for (let i = buckets.length - 1; i >= 0 && result.length < k; i--) {
14 if (buckets[i].length > 0) {
15 result.push(...buckets[i]);
16 }
17 }
18 return result.slice(0, k);
19}
20
21const nums = [1, 1, 1, 2, 2, 3];
22const k = 2;
23console.log(topKFrequent(nums, k));
24In JavaScript, elements are counted via a map and ordered into buckets representing frequencies, making it possible to fetch the most frequent.