old_contents: """Leetcode 394. Decode String
  Medium
  
  URL: https://leetcode.com/problems/decode-string/
  
  Given an encoded string, return its 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
  
          Time complexity: O(m*n), where 
            - m is the length of string, and
            - n is maximun number.
          Space complexity: O(m).
          """
          # Use stack of [string, repeated times] to get decoded string in 1st element.
          stack = [['', 1]]
          num = ''
  
          for c in s:
              if c.isdigit():
                  # Since digit means repeated times, accumulate them.
                  num += c
              elif c == '[':
                  # Start accumulating string with repeated times.
                  stack.append(['', int(num)])
                  num = ''
              elif c == ']':
                  # Stop accumulating string, and put it to the last stack.
                  string, times = stack.pop()
                  stack[-1][0] += string * times
              else:
                  # Continue accumulating string.
                  stack[-1][0] += c
  
          return stack[0][0]
  
  
  def main():
      # s = "3[a]2[bc]", return "aaabcbc".
      s = "3[a]2[bc]"
      print Solution().decodeString(s)
  
      # s = "3[a2[c]]", return "accaccacc".
      s = "3[a2[c]]"
      print Solution().decodeString(s)
  
      # s = "2[abc]3[cd]ef", return "abcabccdcdcdef".
      s = "2[abc]3[cd]ef"
      print Solution().decodeString(s)
  
  
  if __name__ == '__main__':
      main()
  
new_contents: """Leetcode 394. Decode String
  Medium
  
  URL: https://leetcode.com/problems/decode-string/
  
  Given an encoded string, return its 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 SolutionStringRepeatedTimesStackIter(object):
      def decodeString(self, s):
          """
          :type s: str
          :rtype: str
  
          Time complexity: O(m*n), where 
            - m is the length of string, and
            - n is maximun number.
          Space complexity: O(m).
          """
          # Use stack of [string, repeated times] to get decoded string in 1st element.
          stack = [['', 1]]
          num = ''
  
          for c in s:
              if c.isdigit():
                  # Since digit means repeated times, accumulate them.
                  num += c
              elif c == '[':
                  # Start accumulating string with repeated times.
                  stack.append(['', int(num)])
                  num = ''
              elif c == ']':
                  # Stop accumulating string, and put it to the last stack.
                  string, times = stack.pop()
                  stack[-1][0] += string * times
              else:
                  # Continue accumulating string.
                  stack[-1][0] += c
  
          return stack[0][0]
  
  
  def main():
      # s = "3[a]2[bc]", return "aaabcbc".
      s = "3[a]2[bc]"
      print SolutionStringRepeatedTimesStackIter().decodeString(s)
  
      # s = "3[a2[c]]", return "accaccacc".
      s = "3[a2[c]]"
      print SolutionStringRepeatedTimesStackIter().decodeString(s)
  
      # s = "2[abc]3[cd]ef", return "abcabccdcdcdef".
      s = "2[abc]3[cd]ef"
      print SolutionStringRepeatedTimesStackIter().decodeString(s)
  
  
  if __name__ == '__main__':
      main()
  
subject: Revise class name