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).
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.