LeetCode 1207. Unique Number of Occurrences Solution in Java, C++, Python & More | Explanation + Code

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.

PythonJavaC++GoTypeScript

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

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(); } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)