NestedHard.py

def solution(S):
  stack = []
  for sign in S:
    if sign == '(' or sign == '[' or sign =='{':
      stack.append(sign)
    else:
      if len(stack) == 0:
        return 0
      stackLast = stack[-1]
      if sign == ')' and stackLast == '(':
        stack.pop()
      elif sign == ']' and stackLast == '[':
        stack.pop()
      elif sign == '}' and stackLast == '{':
        stack.pop()
      else:
        return 0

  if len(stack) == 0:
    return 1
  else:
    return 0
print (solution('[[]]'))
# A string S consisting of N characters is considered to be properly nested if any of the following conditions is true:
#
# S is empty;
# S has the form "(U)" or "[U]" or "{U}" where U is a properly nested string;
# S has the form "VW" where V and W are properly nested strings.
# For example, the string "{[()()]}" is properly nested but "([)()]" is not.
#
# Write a function:
#
# def solution(S)
#
# that, given a string S consisting of N characters, returns 1 if S is properly nested and 0 otherwise.
#
# For example, given S = "{[()()]}", the function should return 1 and given S = "([)()]", the function should return 0, as explained above.
#
# Assume that:
#
# N is an integer within the range [0..200,000];
# string S consists only of the following characters: "(", "{", "[", "]", "}" and/or ")".
# Complexity:
#
# expected worst-case time complexity is O(N);
# expected worst-case space complexity is O(N) (not counting the storage required for input arguments).