[LeetCode] Unique Number of Occurrences

題目：Unique Number of Occurrences
難度：Easy

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

---

時間複雜度：因要遍歷整個矩陣一次，故為 O(N)，N 為矩陣長度
空間複雜度：因需要一個 dictionary 儲存各數字出現的次數，故為 O(N)，N 為矩陣長度

public class Solution {
    public bool UniqueOccurrences(int[] arr) {
        IDictionary<int, int> dict = new Dictionary<int, int>();

        for (int i = 0; i < arr.Length; i++)
        {
            // 若 dict 尚未有 arr[i]，則新增
            if (dict.ContainsKey(arr[i]) == false)
            {
                dict.Add(arr[i], 1);
            }
            // 若 arr[i] 已存在於 dict，則累加
            else
            {
                dict[arr[i]]++;
            }
        }

        return dict.Keys.Count == dict.Values.ToHashSet().Count;
    }
}