Pop Operation on Stack in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Python

def pop(stack):
    if not stack:
        return None
    return stack[1]()

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

Aappend

Bpop

Cremove

Dinsert

Attempts:

3 left

💡 Hint

Common Mistakes

Using append() which adds an element instead of removing.

Using remove() which removes by value, not by position.

The pop() method removes and returns the last item from the list, which is the top of the stack.

2fill in blank

medium

Complete the code to check if the stack is empty before popping.

DSA Python

def safe_pop(stack):
    if len(stack) [1] 0:
        return stack.pop()
    return None

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

A==

B!=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' which only checks if stack is exactly zero.

Using '<' which checks for less than zero, impossible for length.

We check if the stack length is greater than 0 before popping to avoid errors.

3fill in blank

hard

Fix the error in the pop function to correctly remove the top element.

DSA Python

def pop_element(stack):
    if stack:
        return stack[1]()
    return None

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

Aappend

Bremove

Cpop

Dinsert

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop(0) which removes the bottom element.

Using remove() which removes by value, not position.

Using pop() without arguments removes the last element (top of stack). pop(0) removes the first element, which is incorrect for stack.

4fill in blank

hard

Fill both blanks to create a function that pops the top element and returns the updated stack.

DSA Python

def pop_and_return(stack):
    if stack:
        stack[1]()
    return [2]

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

Apop

Bstack

CNone

Dappend

Attempts:

3 left

💡 Hint

Common Mistakes

Returning None instead of the updated stack.

Using append() which adds elements instead of removing.

Calling pop() removes the top element. Returning stack shows the updated stack.

5fill in blank

hard

Fill all three blanks to pop the top element only if the stack is not empty and return the popped element.

DSA Python

def conditional_pop(stack):
    if len(stack) [1] 0:
        return stack[2]()
    else:
        return [3]

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

A>

Bpop

CNone

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '>' to check stack length.

Returning stack instead of None when empty.

Check if stack length is greater than 0, then pop the top element. Otherwise, return None.

Practice

1. What does the pop() operation do in a stack?

easy

A. Clears all elements from the stack

B. Adds a new element to the bottom of the stack

C. Returns the bottom element without removing it

D. Removes and returns the top element of the stack

Solution

1. Step 1: Understand stack behavior

   A stack follows Last In First Out (LIFO) principle, so the last added element is on top.

2. Step 2: Define pop operation

   The pop() removes and returns this top element, reducing stack size by one.

3. Final Answer:

   Removes and returns the top element of the stack -> Option D

4. Quick Check:

   pop() = remove top element [OK]

Hint: Pop always removes the last added item [OK]

Common Mistakes:

• Thinking pop removes bottom element
• Confusing pop with push operation
• Assuming pop returns the whole stack

2. Which of the following is the correct syntax to pop an element from a stack named stack in Python?

easy

A. stack.pop(0)

B. pop(stack)

C. stack.pop()

D. stack.remove()

Solution

1. Step 1: Recall Python list pop syntax

   In Python, pop() is a method called on the list object to remove the last element.

2. Step 2: Check each option

   stack.pop() uses stack.pop() which is correct. pop(stack) is invalid syntax. stack.pop(0) pops first element, not top of stack. stack.remove() is invalid method.

3. Final Answer:

   stack.pop() -> Option C

4. Quick Check:

   stack.pop() is correct syntax [OK]

Hint: Use stack.pop() to remove top element [OK]

Common Mistakes:

• Using pop with index 0 removes bottom, not top
• Calling pop as a standalone function
• Using remove() which deletes by value, not position

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

stack = [10, 20, 30]
popped = stack.pop()
print(popped)
print(stack)

medium

A. 30
[10, 20]

B. 10
[20, 30]

C. 20
[10, 30]

D. 30
[20, 10]

Solution

1. Step 1: Identify initial stack and pop operation

   Stack starts as [10, 20, 30]. Calling pop() removes last element 30 and returns it.

2. Step 2: Check printed values

   Variable popped is 30. Stack after pop is [10, 20]. So print outputs 30 and then [10, 20].

3. Final Answer:

   30 [10, 20] -> Option A

4. Quick Check:

   pop() returns last element 30 [OK]

Hint: pop() returns last item, stack shrinks [OK]

Common Mistakes:

• Thinking pop removes first element
• Confusing popped value with stack after pop
• Misreading list order after pop

4. Find the error in the following code that tries to pop an element from an empty stack:

stack = []
popped = stack.pop()
print(popped)

medium

A. SyntaxError due to missing parentheses

B. IndexError because pop on empty list is invalid

C. TypeError because pop requires an argument

D. No error, prints null

Solution

1. Step 1: Understand pop on empty list

   Calling pop() on an empty list raises an IndexError because there is no element to remove.

2. Step 2: Check error types

   SyntaxError and TypeError do not apply here. The code does not print null; it raises an error.

3. Final Answer:

   IndexError because pop on empty list is invalid -> Option B

4. Quick Check:

   pop() on empty list = IndexError [OK]

Hint: Pop on empty stack causes IndexError [OK]

Common Mistakes:

• Assuming pop returns null if empty
• Confusing error type with SyntaxError
• Not checking if stack is empty before pop

5. Given a stack implemented as a list, how can you safely pop all elements without causing errors, and print them in order they were removed?

hard

A. Use a while loop: while stack: print(stack.pop())

B. Use a for loop: for i in range(len(stack)): print(stack.pop())

C. Use recursion without base case to pop elements

D. Directly print stack.pop() without checking if empty

Solution

1. Step 1: Understand safe popping

   Using while stack: checks if stack is not empty before popping, avoiding errors.

2. Step 2: Evaluate options

   Use a while loop: while stack: print(stack.pop()) safely pops all elements printing each. Use a for loop: for i in range(len(stack)): print(stack.pop()) uses fixed range which fails if stack size changes during loop. Use recursion without base case to pop elements causes infinite recursion. Directly print stack.pop() without checking if empty risks error if stack empty.

3. Final Answer:

   Use a while loop: while stack: print(stack.pop()) -> Option A

4. Quick Check:

   while stack: pop safely [OK]

Hint: Check stack non-empty before pop in loop [OK]

Common Mistakes:

• Using for loop with changing stack size
• Not checking empty stack before pop
• Infinite recursion without base case