old_contents: """Leetcode 1249. Minimum Remove to Make Valid Parentheses
  Medium
  
  URL: https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
  
  Given a string s of '(' , ')' and lowercase English characters. 
  
  Your task is to remove the minimum number of parentheses ( '(' or ')',
  in any positions ) so that the resulting parentheses string is valid and
  return any valid string.
  
  Formally, a parentheses string is valid if and only if:
  - It is the empty string, contains only lowercase characters, or
  - It can be written as AB (A concatenated with B), where A and B are valid strings, or
  - It can be written as (A), where A is a valid string.
  
  Example 1:
  Input: s = "lee(t(c)o)de)"
  Output: "lee(t(c)o)de"
  Explanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.
  
  Example 2:
  Input: s = "a)b(c)d"
  Output: "ab(c)d"
  
  Example 3:
  Input: s = "))(("
  Output: ""
  Explanation: An empty string is also valid.
  
  Example 4:
  Input: s = "(a(b(c)d)"
  Output: "a(b(c)d)"
  
  Constraints:
  - 1 <= s.length <= 10^5
  - s[i] is one of  '(' , ')' and lowercase English letters.
  """
  
  
  class Solution(object):
      def minRemoveToMakeValid(self, s):
          """
          :type s: str
          :rtype: str
          """
          pass
  
  
  def main():
      pass
  
  
  if __name__ == '__main__':
      main()
  
new_contents: """Leetcode 1249. Minimum Remove to Make Valid Parentheses
  Medium
  
  URL: https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
  
  Given a string s of '(' , ')' and lowercase English characters. 
  
  Your task is to remove the minimum number of parentheses ( '(' or ')',
  in any positions ) so that the resulting parentheses string is valid and
  return any valid string.
  
  Formally, a parentheses string is valid if and only if:
  - It is the empty string, contains only lowercase characters, or
  - It can be written as AB (A concatenated with B), where A and B are valid strings, or
  - It can be written as (A), where A is a valid string.
  
  Example 1:
  Input: s = "lee(t(c)o)de)"
  Output: "lee(t(c)o)de"
  Explanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.
  
  Example 2:
  Input: s = "a)b(c)d"
  Output: "ab(c)d"
  
  Example 3:
  Input: s = "))(("
  Output: ""
  Explanation: An empty string is also valid.
  
  Example 4:
  Input: s = "(a(b(c)d)"
  Output: "a(b(c)d)"
  
  Constraints:
  - 1 <= s.length <= 10^5
  - s[i] is one of  '(' , ')' and lowercase English letters.
  """
  
  
  class SolutionStackAppendValidChars(object):
      def minRemoveToMakeValid(self, s):
          """
          :type s: str
          :rtype: str
  
          Time complexity: O(n).
          Space complexity: O(n).
          """
          # Iterate through chars, add valid chars to valid chars array.
          LOWER_CHARS = 'abcdefghijklmnopqrstuvwxyz'
          stack = []
          valid_chars = []
  
          for c in s:
              if c == '(' or c in LOWER_CHARS:
                  # If char is '(' or lower, add char to valid chars.
                  valid_chars.append(c)
  
                  if c == '(':
                      # If '(', push it to stack.
                      stack.append('(')
  
              elif c == ')' and stack:
                  # If char is ')' and stack contains ')' for matching,
                  # add char to valid chars and pop '('.
                  valid_chars.append(c)
                  stack.pop()
              else:
                  # If not match, ignore ')' and continue.
                  continue
  
          if stack:
              # If there remain '(', remove redundant '(' by its number from backward.
              counter = len(stack)
              for i in range(len(valid_chars) - 1, -1, -1):
                  if counter > 0:
                      if valid_chars[i] == '(':
                          valid_chars[i] = ''
                          counter -= 1
                  else:
                      break
          
          return ''.join(valid_chars)
  
  
  def main():
      # Output: "lee(t(c)o)de"
      s = "lee(t(c)o)de)"
      print SolutionStackAppendValidChars().minRemoveToMakeValid(s)
  
      # Output: "ab(c)d"
      s = "a)b(c)d"
      print SolutionStackAppendValidChars().minRemoveToMakeValid(s)
  
      # Output: ""
      s = "))(("
      print SolutionStackAppendValidChars().minRemoveToMakeValid(s)
  
      # Output: "a(b(c)d)"
      s = "(a(b(c)d)"
      print SolutionStackAppendValidChars().minRemoveToMakeValid(s)
  
      # Output: "()()"
      s = "())()((("
      print SolutionStackAppendValidChars().minRemoveToMakeValid(s)
  
  
  if __name__ == '__main__':
      main()
  
subject: Complete stack append valid char sol w/ time/space complexity