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 <stdio.h>
2#include <stdlib.h>
3#include <string.h>
4
5#define MAX 10000
6
7typedef struct {
8 int value;
9 int count;
10} Freq;
11
12int compare(const void *a, const void *b) {
13 return ((Freq *)b)->count - ((Freq *)a)->count;
14}
15
16int* topKFrequent(int* nums, int numsSize, int k, int* returnSize) {
17 int freqMap[2 * MAX + 1] = {0};
18 Freq freqArray[numsSize];
19 int uniqueCount = 0;
20
21 for(int i = 0; i < numsSize; i++) {
22 freqMap[nums[i] + MAX]++;
23 }
24
25 for(int i = 0; i < 2 * MAX + 1; i++) {
26 if(freqMap[i]) {
27 freqArray[uniqueCount].value = i - MAX;
28 freqArray[uniqueCount].count = freqMap[i];
29 uniqueCount++;
30 }
31 }
32
33 qsort(freqArray, uniqueCount, sizeof(Freq), compare);
34
35 *returnSize = k;
36 int *result = (int*)malloc(sizeof(int) * k);
37 for(int i = 0; i < k; i++) {
38 result[i] = freqArray[i].value;
39 }
40 return result;
41}
42
43int main() {
44 int nums[] = {1, 1, 1, 2, 2, 3};
45 int k = 2;
46 int returnSize;
47 int* result = topKFrequent(nums, 6, k, &returnSize);
48 for(int i = 0; i < returnSize; i++) {
49 printf("%d ", result[i]);
50 }
51 free(result);
52 return 0;
53}
54We use a frequency map to track occurrences of each number. Then, we create a struct array for frequencies, sort it, and return 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.