In this approach, we will first traverse the array to count the occurrences of each element using a hash map. Then, we will use a set to check if these occurrences are unique. If the size of the set matches the size of the occurrence counts, it implies all occurrences are unique.
Time Complexity: O(n), where n is the length of the array.
Space Complexity: O(1), since we're using fixed-size arrays.
1import java.util.HashMap;
2import java.util.HashSet;
3
4public class UniqueOccurrences {
5 public static boolean uniqueOccurrences(int[] arr) {
6 HashMap<Integer, Integer> countMap = new HashMap<>();
7 for (int num : arr) {
8 countMap.put(num, countMap.getOrDefault(num, 0) + 1);
9 }
10 HashSet<Integer> occurrences = new HashSet<>();
11 for (int count : countMap.values()) {
12 if (!occurrences.add(count)) {
13 return false;
14 }
15 }
16 return true;
17 }
18
19 public static void main(String[] args) {
20 int[] arr = {1, 2, 2, 1, 1, 3};
21 System.out.println(uniqueOccurrences(arr));
22 }
23}In the Java implementation, we use a HashMap to track element frequencies and a HashSet to ensure all frequencies are unique.
This alternative approach starts by counting occurrences just like the first one. After that, it stores these counts in a list, sorts the list, and then checks for any consecutive equal elements, which would indicate duplicate occurrences.
Time Complexity: O(n log n), due to sorting.
Space Complexity: O(1), since we work within fixed-sized arrays.
1#include <stdio.h>
2#include <stdbool.h>
3#include <stdlib.h>
4
5int cmpfunc (const void * a, const void * b) {
6 return (*(int*)a - *(int*)b);
7}
8
9bool uniqueOccurrences(int* arr, int arrSize) {
10 int count[2001] = {0};
11 int occ[1001] = {0};
12 int idx = 0;
13
14 for(int i = 0; i < arrSize; i++) {
15 count[arr[i] + 1000]++;
16 }
17
18 for(int i = 0; i < 2001; i++) {
19 if(count[i] > 0) {
20 occ[idx++] = count[i];
21 }
22 }
23
24 qsort(occ, idx, sizeof(int), cmpfunc);
25
26 for(int i = 1; i < idx; i++) {
27 if(occ[i] == occ[i - 1])
28 return false;
29 }
30
31 return true;
32}
33
34int main() {
35 int arr[] = {1, 2, 2, 1, 1, 3};
36 int arrSize = sizeof(arr) / sizeof(arr[0]);
37 printf("%s\n", uniqueOccurrences(arr, arrSize) ? "true" : "false");
38 return 0;
39}In the C implementation using sorting, we use qsort to sort the frequency array and then check if any consecutive elements are equal.