###### tags: `Leetcode` `easy` `hash` `python` `c++`

# 1207. Unique Number of Occurrences

## [題目連結:] https://leetcode.com/problems/unique-number-of-occurrences/description/

## 題目:
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

## 解題想法:
* 此題為，判斷數組中的每個數，出現次數是否皆不同
* 兩個字典判斷:
* 第一個字典
* key: 數字
* val: 出現次數
* 第二個字典
* key: 出現次數
* val: Bool

## Python:

from collections import defaultdict
class Solution(object):
    def uniqueOccurrences(self, arr):
        """
        :type arr: List[int]
        :rtype: bool
        """
        dic=defaultdict(int)
        num=defaultdict(bool)
        for val in arr:
            dic[val]+=1
        for key in dic:
            if dic[key] in num:
                return False
            num[dic[key]]=True
        return True

## C++:

class Solution {
public:
    bool uniqueOccurrences(vector<int>& arr) {
        unordered_map<int,int> dic;
        unordered_map<int,bool> appear;
        for (int val: arr){
            dic[val]+=1;
        }
        //unordered_map
        //first: key
        //second: value
        for (const auto & item: dic){
            if (appear.find(item.second)!=appear.end())
                return false;
            appear[item.second]=true;
        }
        return true;
    }
};