N-Repeated Element in Size 2N Array

You are given an integer array nums with the following properties:

  • nums.length == 2 * n.
  • nums contains n + 1 unique elements.
  • Exactly one element of nums is repeated n times.

Return the element that is repeated n times.

 

Example 1:

Input: nums = [1,2,3,3]
Output: 3

Example 2:

Input: nums = [2,1,2,5,3,2]
Output: 2

Example 3:

Input: nums = [5,1,5,2,5,3,5,4]
Output: 5

 

Constraints:

  • 2 <= n <= 5000
  • nums.length == 2 * n
  • 0 <= nums[i] <= 104
  • nums contains n + 1 unique elements and one of them is repeated exactly n times.
SOLUTION:
class Solution:
    def repeatedNTimes(self, nums: List[int]) -> int:
        n = len(nums)
        nums.sort()
        i = 0
        while i < n:
            ctr = 1
            while i < n - 1 and nums[i] == nums[i + 1]:
                ctr += 1
                i += 1
            i += 1
            if 2 * ctr == n:
                return nums[i - 1]

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