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:
        cnt = Counter(arr)
        return len(set(cnt.values())) == len(cnt)

Java

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();
    }
}

C++

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;
    }
};

Go

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
}

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