In a array A of size 2N, there are N+1 unique elements, and exactly one of these elements is repeated N times.
Return the element repeated N times.
Example 1:
Input: [1,2,3,3]
Output: 3
Example 2:
Input: [2,1,2,5,3,2]
Output: 2
Example 3:
Input: [5,1,5,2,5,3,5,4]
Output: 5
Note:
4 <= A.length <= 10000
0 <= A[i] < 10000
A.length is even
class Solution:
    def repeatedNTimes(self, A: List[int]) -> int:
        nums = {}
        for i in A:
            if i not in list(nums.keys()):
                nums[i] = 1
            else:
                nums[i] += 1
        
        print(nums)
        
        for i in nums:
            if nums[i] == len(A) / 2:
                return i
            

class Solution:
    def repeatedNTimes(self, A: List[int]) -> int:
        nums = {}
        for i in A:
            if i not in list(nums.keys()):
                nums[i] = 1
            else:
                nums[i] += 1
        
        print(nums)
        
        for i in nums:
            if nums[i] == len(A) / 2:
                return i
            
