Given an array of integers arr, write a function that returns true if and only if the number of occurrences of each value in the array is unique.
 
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

Difficulty: introductory

Starter code:

class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        

Solutions:
from collections import defaultdict
class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        occurences = defaultdict(int)
        for i in arr:
            occurences[i] += 1
        for i in occurences.values():
            if list(occurences.values()).count(i) > 1:
                return False
        return True

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


class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        l=list(set(arr))
        for i in range(len(l)):
            l[i]=arr.count(l[i])
        if len(l)==len(set(l)):
            return True
        else:
            return False
        


class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        occur = {}
        check = set()
        for num in arr:
            if num not in occur:
                occur[num] = arr.count(num)
        for num in occur.values():
            check.add(num)
        if len(occur) == len(check):
            return True
        return False

class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        res = False
        occ=[]
        n=0
        for i in range(0,len(arr)):
            if arr[i] in arr[0:i]:
                pass
            else:  
                n = arr.count(arr[i])
                occ.append(n)


        c=0

        for j in range(0,len(occ)):
            for k in range(j+1,len(occ)):
                if occ[j]==occ[k]:
                    c=1
                    res=False
                    break
                else:
                    c=0
                    res=True

            if c==1:
                break
        
        return res


class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        count = {}
        for a in arr:
            if a not in count:
                count[a] = 1
            else:
                count[a] += 1
        occ = []
        for c in list(count.values()):
            if c not in occ:
                occ.append(c)
            else:
                return False
        return True


class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        count = Counter(arr)
        s = set()
        
        for k,v in list(count.items()):
            if v in s:
                return False
            else:
                s.add(v)
        return True
            
            


class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        from collections import Counter
        count_dict = dict(Counter(arr))
        count_occurrences = [j for i,j in list(count_dict.items())]
        return len(set(arr)) == len(set(count_occurrences))


class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        occd = {}
        for n in arr:
            if n in occd:
                occd[n] += 1
            else:
                occd[n] = 1
                
        s = set()
        for k, v in occd.items():
            if v in s:
                return False
            s.add(v)
            
        return True

class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        counter = {}

        for i in arr:
            counter[i] = counter.get(i, 0) + 1

        return len(counter.values()) == len(set(list(counter.values())))

class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        
        dct = {}
        
        for num in arr:
            if num in dct:
                dct[num]+=1
            else:
                dct[num] = 1
                
        return len(list(dct.values())) == len(set(dct.values()))
                
       
            


class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        rec = [arr.count(elem) for elem in set(arr)]
        return len(rec) == len(set(rec))


class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        unique = set(arr)
        sol = []
        for i in unique:
            sol.append(arr.count(i))
        print(sol)
        for i in sol:
            if sol.count(i) > 1:
                return False
        return True