Python solution o(n) no xor


#1

simply use dictionary (mine init in o(n) but you can do it on demand inside the for loop)
then run on array and dell who happen twice

def singleNumber(self, A):
    d=dict.fromkeys(A,0)
    for i in A:
        d[i] += 1
        if (d[i] == 2):
            del d[i]
    return int(list(d)[0])