Pop Operation on Stack in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after popping from the stack?

Consider the following Python code that uses a list as a stack. What will be printed after the pop operation?

DSA Python

stack = [10, 20, 30, 40]
popped = stack.pop()
print(stack)

A [10, 20, 30]

B [10, 20, 30, 40]

C [20, 30, 40]

D []

💡 Hint

Remember, pop() removes the last item added to the list.

The pop() method removes the last element from the list, which is 40. So the remaining stack is [10, 20, 30].

❓ Predict Output

intermediate

What does this pop operation return?

What value is stored in popped after this code runs?

DSA Python

stack = ['a', 'b', 'c']
popped = stack.pop()
print(popped)

A b

B c

C a

D None

💡 Hint

Pop returns the last item removed from the stack.

The last item in the list is 'c', so pop() removes and returns 'c'.

🔧 Debug

advanced

What error does this pop operation cause?

What error will this code raise when executed?

DSA Python

stack = []
popped = stack.pop()

A No error

B KeyError

C IndexError

D TypeError

💡 Hint

Think about what happens when you pop from an empty list.

Popping from an empty list raises an IndexError because there is no element to remove.

🧠 Conceptual

advanced

How many items remain after multiple pops?

Given the stack [5, 10, 15, 20, 25], after performing pop() three times, how many items remain in the stack?

DSA Python

stack = [5, 10, 15, 20, 25]
stack.pop()
stack.pop()
stack.pop()
print(len(stack))

A 2

B 3

C 5

D 0

💡 Hint

Each pop removes one item from the end.

Starting with 5 items, popping 3 removes 3 items, leaving 2 items.

🚀 Application

expert

What is the final stack after this sequence of operations?

Starting with an empty stack, perform these operations in order:

1. push 1
2. push 2
3. pop
4. push 3
5. pop
6. push 4

What is the final state of the stack?

DSA Python

stack = []
stack.append(1)
stack.append(2)
stack.pop()
stack.append(3)
stack.pop()
stack.append(4)
print(stack)

A [1, 2, 3, 4]

B [4]

C [1, 3, 4]

D [1, 4]

💡 Hint

Track each operation step by step.

Operations:
push 1 → [1]
push 2 → [1, 2]
pop → removes 2 → [1]
push 3 → [1, 3]
pop → removes 3 → [1]
push 4 → [1, 4]

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