# Time:  O(n^3)
# Space: O(n)
#
# Additive number is a positive integer whose digits can form additive
# sequence.
#
# A valid additive sequence should contain at least three numbers.
# Except for the first two numbers, each subsequent number in the
# sequence
# must be the sum of the preceding two.
#
# For example:
# "112358" is an additive number because the digits can form an additive
# sequence:
#   1, 1, 2, 3, 5, 8.
#
# 1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8
# "199100199" is also an additive number, the additive sequence is:
# 1, 99, 100, 199.
#
# 1 + 99 = 100, 99 + 100 = 199
# Note: Numbers in the additive sequence cannot have leading zeros,
# so sequence 1, 2, 03 or 1, 02, 3 is invalid.
#
# Given a string represents an integer, write a function to determine
# if it's an additive number.
#
# Follow up:
# How would you handle overflow for very large input integers?
#

try:
    xrange          # Python 2
except NameError:
    xrange = range  # Python 3


class Solution(object):
    def isAdditiveNumber(self, num):
        """
        :type num: str
        :rtype: bool
        """
        def add(a, b):
            res, carry, val = "", 0, 0
            for i in xrange(max(len(a), len(b))):
                val = carry
                if i < len(a):
                    val += int(a[-(i + 1)])
                if i < len(b):
                    val += int(b[-(i + 1)])
                carry, val = val / 10, val % 10
                res += str(val)
            if carry:
                res += str(carry)
            return res[::-1]

        for i in xrange(1, len(num)):
            for j in xrange(i + 1, len(num)):
                s1, s2 = num[0:i], num[i:j]
                if (len(s1) > 1 and s1[0] == '0') or \
                   (len(s2) > 1 and s2[0] == '0'):
                    continue

                expected = add(s1, s2)
                cur = s1 + s2 + expected
                while len(cur) < len(num):
                    s1, s2, expected = s2, expected, add(s2, expected)
                    cur += expected
                if cur == num:
                    return True
        return False