Please use Python for this question: 1. Inheritance (based on 8.38) You MUST use inheritance for this problem. A stack is a sequence container type that, like a queue, supports very restrictive access methods: all insertions and removals are from one end of the stack, typically referred to as the top of the stack. A stack is often referred to as a last-in first-out (LIFO) container because the last item inserted is the first removed. Implement a Stack class using inheritance. Note that this means you may be able to inherit some of the methods below: a. Constructor/_init__ - Can construct either an empty stack, or initialized with a list of items, the first item is at the bottom, the last is at the top. b. push() – take an item as input and push it on the top of the stack c. pop() – remove and return the item at the top of the stack d. isEmpty() – returns True if the stack is empty, False otherwise e. [] – return the item at a given location, [0] is at the bottom of the stack f. len() – return length of the stack Output for this question: ''' >>> s = Stack() >>> s.push('apple') >>> s Stack(['apple']) >>> s.push('pear') >>> s.push('kiwi') >>> s Stack(['apple', 'pear', 'kiwi']) >>> top = s.pop() >>> top 'kiwi' >>> s Stack(['apple', 'pear']) >>> len(s) 2 >>> s.isEmpty() False >>> s.pop() 'pear' >>> s.pop() 'apple' >>> s.isEmpty() True >>> s = Stack(['apple', 'pear', 'kiwi']) >>> s = Stack(['apple', 'pear', 'kiwi']) >>> s[0] 'apple' >>> '''

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter18: Stacks And Queues

Section: Chapter Questions

Problem 3PE

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Please use Python for this question:

1. Inheritance (based on 8.38) You MUST use inheritance for this problem. A stack is a sequence container type that, like a queue, supports very restrictive access methods: all insertions and removals are from one end of the stack, typically referred to as the top of the stack. A stack is often referred to as a last-in first-out (LIFO) container because the last item inserted is the first removed. Implement a Stack class using inheritance. Note that this means you may be able to inherit some of the methods below:

a. Constructor/_init__ - Can construct either an empty stack, or initialized with a list of items, the first item is at the bottom, the last is at the top.

b. push() – take an item as input and push it on the top of the stack

c. pop() – remove and return the item at the top of the stack

d. isEmpty() – returns True if the stack is empty, False otherwise

e. [] – return the item at a given location, [0] is at the bottom of the stack

f. len() – return length of the stack

Output for this question:

'''

>>> s = Stack()

>>> s.push('apple')

>>> s

Stack(['apple'])

>>> s.push('pear')

>>> s.push('kiwi')

>>> s

Stack(['apple', 'pear', 'kiwi'])

>>> top = s.pop()

>>> top

'kiwi'

>>> s

Stack(['apple', 'pear'])

>>> len(s)

>>> s.isEmpty()

False

>>> s.pop()

'pear'

>>> s.pop()

'apple'

>>> s.isEmpty()

True

>>> s = Stack(['apple', 'pear', 'kiwi'])

>>> s[0]

'apple'

>>>

'''

2. 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.

output for this question:

>>> 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

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:

a. If the character is one of opening marks(,[,{ push it on the stack.

b. 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.

c. 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.

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