Please answer the question below in Python: 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 Your solution must use a Stack. For, example: >>> 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 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: Create an empty stack. Iterate over the characters in the given string: If the character is one of opening marks(,[,{ push it on the stack. 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. 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. Once the iteration is finished, you know that the parentheses match if and only if the stack is empty.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Please answer the question below in Python:

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 Your solution must use a Stack. For, example:

>>> 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

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:

Create an empty stack.
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.
Once the iteration is finished, you know that the parentheses match if and only if the stack is empty.

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