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.
1using System;
2using System.Collections.Generic;
3using System.Linq;
4
5public class Program {
6 public static bool UniqueOccurrences(int[] arr) {
7 Dictionary<int, int> countMap = new Dictionary<int, int>();
8 foreach (int num in arr) {
9 if (countMap.ContainsKey(num)) countMap[num]++;
10 else countMap[num] = 1;
11 }
12 List<int> occurrences = countMap.Values.ToList();
13 occurrences.Sort();
14 for (int i = 1; i < occurrences.Count; i++) {
15 if (occurrences[i] == occurrences[i - 1]) return false;
16 }
17 return true;
18 }
19
20 public static void Main() {
21 int[] arr = {1, 2, 2, 1, 1, 3};
22 Console.WriteLine(UniqueOccurrences(arr));
23 }
24}In C#, we use Linq to turn the map of occurrences into a list, sort it, and check if there are any duplicates in sorted order.