XOR BIT 136 single number

#O(n) space by using hash table
    '''
    def singleNumber(self, nums):
        dict_list = dict()
        for i in nums:
            try:
                dict_list[i].pop()
                del dict_list[i]
            except:
                dict_list[i] = [1]
        for a,b in dict_list.items():
            return a
    
    #O(n) space 2*(a+b+c) - (a+a+b+b+c) = c where c is the single one we are looking for
    def singleNumber(self, nums):
        return 2*sum(set(nums)) - sum(nums)
    '''
    #O(1) space, use XOR ops towards each element since a^0 = a; a^a = 0; a^a^b = b
    def singleNumber(self, nums):
        res = nums[0]
        for i in range(1,len(nums)):
            res ^= nums[i]
        return res