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+# Time:  O(loga)%0A# Space: O(1)%0A%0A# Given a positive integer a,%0A# find the smallest positive integer b whose multiplication of each digit equals to a.%0A#%0A# If there is no answer or the answer is not fit in 32-bit signed integer, then return 0.%0A#%0A# Example 1%0A# Input:%0A#%0A# 48 %0A# Output:%0A# 68%0A# Example 2%0A# Input:%0A#%0A# 15%0A# Output:%0A# 35%0A%0Aclass Solution(object):%0A    def smallestFactorization(self, a):%0A        %22%22%22%0A        :type a: int%0A        :rtype: int%0A        %22%22%22%0A        if a %3C 2:%0A            return a%0A        result, mul = 0, 1%0A        for i in reversed(xrange(2, 10)):%0A            while a %25 i == 0:%0A                a /= i%0A                result = mul*i + result%0A                mul *= 10%0A        return  result if a == 1 and result %3C 2**31 else 0%0A  %0A

Create minimum-factorization.py