Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
Example 1:
Input: nums = [2,2,1]
Output: 1
Example 2:
Input: nums = [4,1,2,1,2]
Output: 4
Example 3:
Input: nums = [1]
Output: 1
Solution
from collections import Counter
class Solution:
def singleNumber(self, nums: List[int]) -> int:
c = Counter(nums)
for i, j in c.items():
if j == 1:
return i