README_EN.md

1207. Unique Number of Occurrences

中文文档

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

Python3

class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        counter = Counter(arr)
        s = set()
        for num in counter.values():
            if num in s:
                return False
            s.add(num)
        return True

Java

class Solution {
    public boolean uniqueOccurrences(int[] arr) {
        Map<Integer, Integer> counter = new HashMap<>();
        for (int e : arr) {
            counter.put(e, counter.getOrDefault(e, 0) + 1);
        }
        Set<Integer> s = new HashSet<>();
        for (int num : counter.values()) {
            if (s.contains(num)) {
                return false;
            }
            s.add(num);
        }
        return true;
    }
}

C++

class Solution {
public:
    bool uniqueOccurrences(vector<int>& arr) {
        unordered_map<int, int> counter;
        for (auto e : arr) {
            ++counter[e];
        }
        unordered_set<int> s;
        for (auto e : counter) {
            int num = e.second;
            if (s.count(num)) return false;
            s.insert(num);
        }
        return true;
    }
};

Go

func uniqueOccurrences(arr []int) bool {
	counter := make(map[int]int)
	for _, e := range arr {
		counter[e]++
	}
	s := make(map[int]bool)
	for _, num := range counter {
		if s[num] {
			return false
		}
		s[num] = true
	}
	return true
}

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