LeetCode 1207. Unique Number of Occurrences

原题链接在这里：https://leetcode.com/problems/unique-number-of-occurrences/

题目：

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

Constraints:

• 1 <= arr.length <= 1000
• -1000 <= arr[i] <= 1000

题解：

Have a map count num to its frequency.

Then go through the map frequency and use a set to record if there is any duplicate.

Time Complexity: O(n). n = arr.length.

Space: O(n).

AC Java:

class Solution {
    public boolean uniqueOccurrences(int[] arr) {
        if(arr == null || arr.length == 0){
            return true;
        }
        
        HashMap<Integer, Integer> numToFreq = new HashMap<>();
        HashSet<Integer> freqSet = new HashSet<>();
        for(int num : arr){
            numToFreq.put(num, numToFreq.getOrDefault(num, 0) + 1);
        }
        
        for(int n : numToFreq.values()){
            if(freqSet.contains(n)){
                return false;
            }
            
            freqSet.add(n);
        }
        
        return true;
    }
}

 AC C++:

class Solution {
public:
    bool uniqueOccurrences(vector<int>& arr) {
        unordered_map<int, int> map;
        unordered_set<int> set;

        for(auto& num : arr){
            map[num]++;
        }

        for(auto& kv : map){
            if(!set.insert(kv.second).second){
                return false;
            }
        }

        return true;
    }
};

AC Python:

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