#Add Binary

#Difficulty:Easy

#

#Given two binary strings, return their sum (also a binary string).

#

#The input strings are both non-empty and contains only characters 1 or 0.

#

#Example 1:

#

#Input: a = "11", b = "1"

#Output: "100"

#Example 2:

#

#Input: a = "1010", b = "1011"

#Output: "10101"

class AddBinary(object):

def __init__(self):

self.answer = ""

#if one number is short then pad it with zeros by the difference of the 2 numbers

def padWithZeros(self, num, numberOfZeros):

for i in range(numberOfZeros):

num = "0"+num

return num

def xor(self, str1, str2):

return str(int(bool(int(str1)) ^ bool(int(str2))))

def xnor(self, str1, str2):

xor_answer = self.xor(str1, str2)

return str(int(not int(xor_answer)))

def addition(self, num1, num2, carry, i1, i2):

if(i1 < 0 and i2 < 0):

self.answer = carry+self.answer

return

binary1 = num1[i1]

binary2 = num2[i2]

i1 = i1-1

i2 = i2-1

if carry == "0":

self.answer = self.xor(binary1, binary2)+self.answer

if(binary1 == "1" and binary2 == "1"):

carry = "1"

else:

#if carry = 1

self.answer = self.xnor(binary1, binary2)+self.answer

if(binary1 == "0" and binary2 == "0"):

carry = "0"

self.addition(num1, num2, carry, i1, i2)

def solution(self, num1, num2):

if(len(num1) < len(num2)):

num1 = self.padWithZeros(num1, len(num2)-len(num1))

elif(len(num2) < len(num1)):

num2 = self.padWithZeros(num2, len(num1)-len(num2))

self.addition(num1, num2, "0", len(num1)-1, len(num2)-1)

print self.answer

#Main

o = AddBinary()

o.solution("1010", "1011")

o2 = AddBinary()

o2.solution("11", "11")