Simplest iterative python using same logic as adding binary numbers by hand

• I don't like writing concise or obfuscated code. It makes it hard to read, understand and maintain. Could be done in lesser number of lines. Comments?

``````class Solution(object):
def addBinary(self, a, b):
"""
:type a: str
:type b: str
:rtype: str
"""
# mapping of sum of three binary numbers - carry from prev sum, bit of one number, bit of second number
dic = {"000": ("0","0") ,
"001": ("0","1"),
"010": ("0","1"),
"011": ("1", "0"),
"100": ("0","1"),
"101": ("1","0"),
"110": ("1", "0"),
"111": ("1","1")}

mins,maxs = (a,b) if len(a) < len(b) else (b,a)

carry = "0"
ans   = ""
i = len(mins) - 1
j = len(maxs) - 1

while i >= 0: # has something in small string to add
carry, x = dic[mins[i] + maxs[j] + carry]
ans += x
i -= 1
j -= 1

while j >= 0: # has something in large string to add
carry, x = dic["0" + maxs[j] + carry] # "0" replaces max[i], we ran out of bits in small string
ans += x
j -= 1

if carry == "1": # got carry, append it
ans += carry

return ans[::-1] # reverse result

# one liner python
# return bin(int(a,2) + int(b,2))[2:]``````

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