Question

1. Write a client function parenthesesMatch that given a string containing only the characters for parentheses, braces or curly braces, i.e., the characters in ’([{}])’, returns True if the parentheses, brackets and braces match and False otherwise. Your solution must use a Stack. For, example:

>>> parenthesesMatch('(){}[]')

True

>>> parenthesesMatch('{[()]}')

True

>>> parenthesesMatch('((())){[()]}')

True

>>> parenthesesMatch('(}')

False

>>> parenthesesMatch('({])')

False

>>> parenthesesMatch('((())')

False

>>> parenthesesMatch('(()))')

False

>>> Hint: It is not sufficient to just count the number of opening and closing marks. But, it is easy to write this as a simple application of the Stack class.   Here is an algorithm:

1. Create an empty stack.
2. Iterate over the characters in the given string:
   1. If the character is one of opening marks(,[,{ push it on the stack.
   2. If the character is one of the closing marks ),],} and the stack is empty, then there were not enough preceding opening marks, so return False.
   3. If the character is a closing mark and the stack is not empty, pop an (opening) mark from the stack. If they are not of the same type, ie., ( and ) or [ and ] or { and }, return False, if they are of the same type, move on to the next char.
3. Once the iteration is finished, you know that the parentheses match if and only if the stack is empty.

use python3.7

check your answer

>>> parenthesesMatch('(){}[]')
True
>>> parenthesesMatch('{[()]}')
True
>>> parenthesesMatch('((())){[()]}')
True
>>> parenthesesMatch('(}')
False
>>> parenthesesMatch(')(][') # right number, but out of order
False
>>> parenthesesMatch('([)]') # right number, but out of order
False
>>> parenthesesMatch('({])')
False
>>> parenthesesMatch('((())')
False
>>> parenthesesMatch('(()))')
False

Answer 1

CODE:

#method to check valid parentheses
def parenthesesMatch(str):

#initializing empty stack
stack = [];

#iterting string
for i in range(0,len(str)):

#if chracter is opening bracket
if(str[i]=='(' or str[i]=='{' or str[i]=='[' ):
#push on stack
stack.append(str[i])

#if character is closing bracket
else:
#if stack is empty
if(len(stack)<1):
#invalid parantheses
return False

#if string character is ')'
if(str[i]==')'):

#top of stack is not '('
if(stack[len(stack)-1]!='('):

#invalid parantheses
return False;

#removing top of stack
stack.pop();

#if string character is '}'   
if(str[i]=='}'):
  
#top of stack is not '{'
if(stack[len(stack)-1]!='{'):

#invalid parantheses
return False;

#removing top of stack
stack.pop();

#if string character is ']'   
if(str[i]==']'):
  
#top of stack is not '['
if(stack[len(stack)-1]!='['):
#invalid parantheses
return False;

#removing top of stack
stack.pop();

# if stack is not empty
if(len(stack)!=0):
# invalid parantheses
return False;

#otherwise valid parantheses
return True;
INDENTED CODE SCREENSHOT:

OUTPUT: