Find single elements among duplicates

forever_alone

a = [7, 2, 2, 3, 3, 4, 4, 55, 55, 100, 100]

b = 0

#for i in range(len(a)):

#you will find the forever alone element in the array


def XOR():

    global b
    for num in a:    #use XOR to iterate over the do the solution

        print "b = %d" %b
        print "num = %d" %num  #o(n) complexity
        b ^= num
        print "result = %d" %b
        print '\n'
XOR()


Logic 

A number xor'd with itself is 0. So the first time you see a value and xor it, it sets the bits and the second time you see it it flips the bits back. So the one that wasn't in a pair never gets flipped back and you end up with that number.