# Python Solution Beats 100%

• Here's how this works:

Go through all the digits from "zero" to "nine", and find a unique character in each that no other digits have. So, for "zero", no other digits have a "z" so you can tell how many "zero"s there are in the string. For "one", there are no unique characters so skip this (we'll get back to this later). For "two", the unique character would be "w"... and so on

Now that you know the number of first pass digits, neglect the first pass digits, do a second pass and find the unique characters that no other remaining digit has. Here, you can count the number of second pass digits in the event that you theoretically removed the first pass digits from the string... Continue on like so.

Now, lets take a second pass digit like 5. You can get a count of 5 because you know the count of the first pass digit 4. How? "f" only occurs in both 4 and 5. You know the number of 4s in the string because its in your first pass (count the number of "u"s in the string). So all you have to do, is count the number of "f"s in the string, and calculate number of 5s equals number of "f"s - number of 4s.

``````    def originalDigits(self, s):
"""
:type s: str
:rtype: str
"""
digits = [0] * 10
# first pass digits
digits[0] = s.count('z')
digits[2] = s.count('w')
digits[4] = s.count('u')
digits[6] = s.count('x')
digits[8] = s.count('g')
# second pass digits
digits[1] = s.count('o') - digits[0] - digits[2] - digits[4]
digits[3] = s.count('h') - digits[8]
digits[5] = s.count('f') - digits[4]
digits[7] = s.count('s') - digits[6]
# third pass digit
digits[9] = s.count('i') - digits[5] - digits[6] - digits[8]

number = ""
for index in xrange(len(digits)):
count = digits[index]
if count > 0:
number += count*str(index)
return number
``````

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