Why Stack Exists and What Problems It Solves in DSA Python - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to push an element onto the stack.

DSA Python

stack = []
stack.[1](5)
print(stack)

Drag options to blanks, or click blank then click option'

Ainsert

Bappend

Cremove

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of append

Using remove which deletes a value

To add an element to the top of the stack, we use the append method.

2fill in blank

medium

Complete the code to remove the top element from the stack.

DSA Python

stack = [1, 2, 3]
top = stack.[1]()
print(top)
print(stack)

Drag options to blanks, or click blank then click option'

Aappend

Binsert

Cremove

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using append instead of pop

Using remove which deletes by value

The pop method removes and returns the last element, which is the top of the stack.

3fill in blank

hard

Fix the error in the code to check if the stack is empty.

DSA Python

stack = []
if len(stack) [1] 0:
    print('Stack is empty')
else:
    print('Stack is not empty')

Drag options to blanks, or click blank then click option'

A==

B!=

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using != which checks for not empty

Using > or < which are incorrect here

To check if the stack is empty, we compare its length to zero using ==.

4fill in blank

hard

Fill both blanks to create a stack and push an element.

DSA Python

stack = [1]
stack.[2](10)
print(stack)

Drag options to blanks, or click blank then click option'

A[]

Bappend

Cpop

D{}

Attempts:

3 left

💡 Hint

Common Mistakes

Using {} which creates a dictionary

Using pop to add elements

We create an empty list with [] and add an element using append.

5fill in blank

hard

Fill all three blanks to pop an element and check if stack is empty.

DSA Python

stack = [5, 6, 7]
top = stack.[1]()
if len(stack) [2] 0:
    print('Stack is empty after popping')
else:
    print('Stack still has elements')
print('Popped element:', [3])

Drag options to blanks, or click blank then click option'

Aappend

Bpop

C==

Dtop

Attempts:

3 left

💡 Hint

Common Mistakes

Using append instead of pop

Using != instead of ==

Printing stack instead of popped element

Use pop to remove the top element, check if length equals zero, and print the popped element stored in top.

Practice

1. Why do we use a stack data structure in programming?

easy

A. To sort numbers automatically

B. To keep track of function calls and return in the correct order

C. To store data in random order for quick access

D. To store data permanently on disk

Solution

1. Step 1: Understand the nature of stack

   A stack works on Last In First Out (LIFO) principle, meaning the last item added is the first to be removed.

2. Step 2: Relate stack to function calls

   Function calls are nested and must return in reverse order of calling, which matches LIFO behavior of stack.

3. Final Answer:

   To keep track of function calls and return in the correct order -> Option B

4. Quick Check:

   Stack = Function call tracking [OK]

Hint: Stack follows LIFO, perfect for tracking nested calls [OK]

Common Mistakes:

• Thinking stack sorts data automatically
• Confusing stack with queue (FIFO)
• Assuming stack stores data permanently

2. Which of the following is the correct way to push an element 'x' onto a stack named stack in Python?

easy

A. stack.push(x)

B. stack.add(x)

C. stack.append(x)

D. stack.insert(x)

Solution

1. Step 1: Recall Python list methods for stack

   Python uses list to implement stack; append() adds element to the top.

2. Step 2: Check other options

   push() and add() are not list methods; insert() inserts at specific index, not top.

3. Final Answer:

   stack.append(x) -> Option C

4. Quick Check:

   Use append() to push in Python stack [OK]

Hint: Use append() to add on top of Python list stack [OK]

Common Mistakes:

• Using push() which is not a Python list method
• Using insert() without index argument
• Confusing add() with append()

3. What will be the output of the following Python code?

stack = []
stack.append(1)
stack.append(2)
stack.append(3)
print(stack.pop())
print(stack)

medium

A. 3
[1, 2]

B. 1
[2, 3]

C. 2
[1, 3]

D. 3
[3, 2, 1]

Solution

1. Step 1: Trace stack after append operations

   After appending 1, 2, 3, stack = [1, 2, 3]

2. Step 2: Understand pop() behavior

   pop() removes and returns last element, so pop() returns 3 and stack becomes [1, 2]

3. Final Answer:

   3 [1, 2] -> Option A

4. Quick Check:

   pop() returns last item, stack updates [OK]

Hint: pop() removes last added element in stack [OK]

Common Mistakes:

• Thinking pop() removes first element
• Confusing append() with insert()
• Expecting stack to be unchanged after pop()

4. The following code is intended to reverse a list using a stack, but it has an error. What is the error?

def reverse_list(lst):
    stack = []
    for item in lst:
        stack.push(item)
    reversed_lst = []
    while stack:
        reversed_lst.append(stack.pop())
    return reversed_lst

print(reverse_list([1, 2, 3]))

medium

A. Using push() method which does not exist for Python lists

B. Using pop() incorrectly without parentheses

C. Not initializing reversed_lst before use

D. Loop condition should be 'while not stack'

Solution

1. Step 1: Check stack operations in Python

   Python lists do not have push() method; append() should be used to add elements.

2. Step 2: Verify other parts of code

   pop() is used correctly with parentheses; reversed_lst is initialized; while stack is correct to check non-empty list.

3. Final Answer:

   Using push() method which does not exist for Python lists -> Option A

4. Quick Check:

   Use append() not push() for Python list stack [OK]

Hint: Python list uses append(), not push() [OK]

Common Mistakes:

• Using push() instead of append()
• Forgetting parentheses on pop()
• Incorrect while loop condition

5. You want to check if a string has balanced parentheses using a stack. Which approach correctly solves this problem?

hard

A. Use a queue to enqueue '(' and dequeue when ')' found

B. Count total '(' and ')' characters; if counts equal, parentheses are balanced

C. Replace all '(' with ')' and check if string is palindrome

D. Push '(' onto stack; pop when ')' found; if stack empty at end, parentheses are balanced

Solution

1. Step 1: Understand balanced parentheses logic

   Each '(' must be matched by a ')' in correct order; stack helps track unmatched '('.

2. Step 2: Explain stack approach

   Push '(' onto stack; on encountering ')', pop stack; if stack empty at end, parentheses are balanced.

3. Step 3: Evaluate other options

   Counting characters ignores order; palindrome check is unrelated; queue (FIFO) does not match nesting order.

4. Final Answer:

   Push '(' onto stack; pop when ')' found; if stack empty at end, parentheses are balanced -> Option D

5. Quick Check:

   Stack tracks open parentheses for balance [OK]

Hint: Use stack push/pop to match parentheses pairs [OK]

Common Mistakes:

• Only counting parentheses without order check
• Using queue instead of stack for nesting
• Ignoring unmatched parentheses inside string