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).
1#include <stdio.h>
2#include <stdlib.h>
3#include <string.h>
4
5#define MAX 10000
6
7typedef struct FreqElem {
8 int value;
9 int count;
10} FreqElem;
11
12int compare(const void *a, const void *b) {
13 return ((FreqElem *)b)->count - ((FreqElem *)a)->count;
14}
15
16int *topKFrequent(int *nums, int numsSize, int k, int *returnSize) {
17 FreqElem buckets[numsSize + 1];
18 memset(buckets, 0, sizeof(buckets));
19
20 for (int i = 0; i < numsSize; ++i) {
21 int found = 0;
22 for (int j = 0; j < numsSize; ++j) {
23 if (buckets[j].count == 0) break;
24 if (buckets[j].value == nums[i]) {
25 buckets[j].count++;
26 found = 1;
27 break;
28 }
29 }
30 if (!found) {
31 for (int j = 0; j < numsSize; ++j) {
32 if (buckets[j].count == 0) {
33 buckets[j].value = nums[i];
34 buckets[j].count = 1;
35 break;
36 }
37 }
38 }
39 }
40
41 qsort(buckets, numsSize, sizeof(FreqElem), compare);
42
43 int *result = (int *)malloc(k * sizeof(int));
44 for (int i = 0; i < k; ++i) {
45 result[i] = buckets[i].value;
46 }
47
48 *returnSize = k;
49 return result;
50}
51
52int main() {
53 int nums[] = {1, 1, 1, 2, 2, 3};
54 int k = 2;
55 int returnSize;
56 int *result = topKFrequent(nums, 6, k, &returnSize);
57 for (int i = 0; i < returnSize; i++) {
58 printf("%d ", result[i]);
59 }
60 free(result);
61 return 0;
62}
63We manually record each element's frequency and sort the list based on counts into a frequency bucket. Then, we retrieve the top k elements.