# 1207. Unique Number of Occurrences

## Description

Given an array of integers arr, return true if the number of occurrences of each value in the array is unique or false otherwise.

Example 1:

Input: arr = [1,2,2,1,1,3]
Output: true
Explanation: The value 1 has 3 occurrences, 2 has 2 and 3 has 1. No two values have the same number of occurrences.

Example 2:

Input: arr = [1,2]
Output: false


Example 3:

Input: arr = [-3,0,1,-3,1,1,1,-3,10,0]
Output: true


Constraints:

• 1 <= arr.length <= 1000
• -1000 <= arr[i] <= 1000

## Solutions

Solution 1: Hash Table

We use a hash table $cnt$ to count the frequency of each number in the array $arr$, and then use another hash table $vis$ to count the types of frequencies. Finally, we check whether the sizes of $cnt$ and $vis$ are equal.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $arr$.

• class Solution {
public boolean uniqueOccurrences(int[] arr) {
Map<Integer, Integer> cnt = new HashMap<>();
for (int x : arr) {
cnt.merge(x, 1, Integer::sum);
}
return new HashSet<>(cnt.values()).size() == cnt.size();
}
}

• class Solution {
public:
bool uniqueOccurrences(vector<int>& arr) {
unordered_map<int, int> cnt;
for (int& x : arr) {
++cnt[x];
}
unordered_set<int> vis;
for (auto& [_, v] : cnt) {
if (vis.count(v)) {
return false;
}
vis.insert(v);
}
return true;
}
};

• class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
cnt = Counter(arr)
return len(set(cnt.values())) == len(cnt)


• func uniqueOccurrences(arr []int) bool {
cnt := map[int]int{}
for _, x := range arr {
cnt[x]++
}
vis := map[int]bool{}
for _, v := range cnt {
if vis[v] {
return false
}
vis[v] = true
}
return true
}

• function uniqueOccurrences(arr: number[]): boolean {
const cnt: Map<number, number> = new Map();
for (const x of arr) {
cnt.set(x, (cnt.get(x) || 0) + 1);
}
return cnt.size === new Set(cnt.values()).size;
}