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.
1#include <iostream>
2#include <vector>
3#include <unordered_map>
4#include <unordered_set>
5
6bool uniqueOccurrences(std::vector<int>& arr) {
7 std::unordered_map<int, int> countMap;
8 for (int num : arr) {
9 ++countMap[num];
10 }
11 std::unordered_set<int> occurrences;
12 for (const auto& pair : countMap) {
13 if (!occurrences.insert(pair.second).second) {
14 return false;
15 }
16 }
17 return true;
18}
19
20int main() {
21 std::vector<int> arr = {1, 2, 2, 1, 1, 3};
22 std::cout << (uniqueOccurrences(arr) ? "true" : "false") << std::endl;
23 return 0;
24}In the C++ version, we use an unordered map to count occurrences of each number, followed by a set to check if any occurrence count is duplicated.
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.
1function uniqueOccurrences(arr) {
2 const countMap = new Map();
3 for (const num of arr) {
4 countMap.set(num, (countMap.get(num) || 0) + 1);
5 }
6 const occurrences = Array.from(countMap.values()).sort((a, b) => a - b);
7 for (let i = 1; i < occurrences.length; i++) {
8 if (occurrences[i] === occurrences[i - 1]) {
9 return false;
10 }
11 }
12 return true;
13}
14
15const arr = [1, 2, 2, 1, 1, 3];
16console.log(uniqueOccurrences(arr));The JavaScript sorting approach involves sorting the list of occurrence counts and checking for any duplicates in consecutive elements.