# Time:  O(n)
# Space: O(h), h is the depth of the recursion

# Given an encoded string, return it's decoded string.
#
# The encoding rule is: k[encoded_string],
# where the encoded_string inside the square brackets is
# being repeated exactly k times. Note that k is guaranteed
# to be a positive integer.
#
# You may assume that the input string is always valid;
# No extra white spaces, square brackets are well-formed, etc.
#
# Furthermore, you may assume that the original data does not
# contain any digits and that digits are only for those repeat numbers, k.
# For example, there won't be input like 3a or 2[4].
#
# Examples:
#
# s = "3[a]2[bc]", return "aaabcbc".
# s = "3[a2[c]]", return "accaccacc".
# s = "2[abc]3[cd]ef", return "abcabccdcdcdef".

class Solution(object):
    def decodeString(self, s):
        """
        :type s: str
        :rtype: str
        """
        curr, nums, strs = [], [], []
        n = 0

        for c in s:
            if c.isdigit():
                n = n * 10 + ord(c) - ord('0')
            elif c == '[':
                nums.append(n)
                n = 0
                strs.append(curr)
                curr = []
            elif c == ']':
                strs[-1].extend(curr * nums.pop())
                curr = strs.pop()
            else:
                curr.append(c)

        return "".join(strs[-1]) if strs else "".join(curr)