1207.Unique Number of Occurrences
===

###### tags: `Easy`,`Array`,`Hash Table`

[1207. Unique Number of Occurrences](https://leetcode.com/problems/unique-number-of-occurrences/)

### 題目描述

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

### 範例

**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`

### 解答

#### Java

class Solution {
    public boolean uniqueOccurrences(int[] arr) {
        HashMap<Integer, Integer> hm = new HashMap<>();
        
        for(Integer num : arr) {
            hm.put(num, hm.getOrDefault(num, 0) + 1);
        }
        
        return hm.size() == new HashSet<>(hm.values()).size();
    }
}

> [name=Kobe][time=Wed, Nov 30]

#### Python

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

> [name=Kobe][time=Wed, Nov 30]

class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        mapping = {}
        for v in arr:
            if v not in mapping:
                mapping[v] = 0
            mapping[v] += 1
        
        value_set = set()
        for v in mapping:
            if mapping[v]  in value_set:
                return False
            value_set.add(mapping[v])

        return True

> 我是廢物寫好長，但是過ㄌ[name=GP]

#### Javascript

function uniqueOccurrences(arr) {
  const map = new Map();
  for (const num of arr) {
    map.set(num, (map.get(num) || 0) + 1);
  }
  return map.size === new Set(map.values()).size;
}

> 昨天學會用map跟set真的是太好用惹！
> [name=Marsgoat] [time= Nov 30, 2022]

#### TypeScript

function uniqueOccurrences(arr: number[]): boolean {
  const map = new Map();
  arr.forEach((key) => {
    map.set(key, (map.get(key) ?? 0) + 1);
  });
  // map.size 跟 value 組成的 set.size 不一樣就代表有重複
  return map.size === new Set(map.values()).size;
}
/* 
 * Map.prototype.set(key, value)：
 *   設置鍵名 key 對應的鍵值為 value，然後回傳整個 Map 結構。
 *   如果 key 已經有值，則鍵值會被更新，否則就新生成該鍵。
 * 
 * Map.prototype.get(key)：
 *   讀取 key 對應的鍵值，如果找不到 key，回傳 undefined。
 */

> the 抄
> [name=sheep] [time= Nov 30, 2022]

### Reference

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