Pop Operation on Stack in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the pop operation do in a stack?

It removes the top item from the stack and returns it, following the Last In First Out (LIFO) principle.

beginner

What happens if you try to pop from an empty stack?

It causes an underflow error because there is no item to remove from the stack.

beginner

In Python, which list method is commonly used to implement the pop operation on a stack?

The pop() method removes and returns the last item of the list, simulating the stack's pop operation.

intermediate

Why is the pop operation considered destructive on a stack?

Because it removes the top element permanently from the stack, changing its state.

beginner

Explain the order of elements after a pop operation on stack with elements: 1 -> 2 -> 3 -> null (3 is top).

After popping, element 3 is removed. The stack becomes: 1 -> 2 -> null, with 2 as the new top.

What does the pop operation return when performed on a stack?

A The top element of the stack

B The bottom element of the stack

C The size of the stack

D Nothing, it only removes the element

Pop removes and returns the top element of the stack.

What is the state of the stack after popping from a stack with elements 5 -> 10 -> 15 -> null (15 is top)?

A 5 -> 10 -> null

B 10 -> 15 -> null

C 5 -> 15 -> null

D Stack becomes empty

Popping removes the top element 15, leaving 5 -> 10 -> null.

Which error occurs if you pop from an empty stack?

A Overflow error

B Type error

C Syntax error

D Underflow error

Underflow error happens when popping from an empty stack.

In Python, which method simulates the pop operation on a stack implemented as a list?

A append()

B remove()

C pop()

D insert()

The pop() method removes and returns the last item, simulating stack pop.

What principle does the pop operation on a stack follow?

A First In First Out (FIFO)

B Last In First Out (LIFO)

C Random Access

D Priority Based

Stacks follow Last In First Out (LIFO) principle.

Describe the pop operation on a stack and what happens to the stack after popping.

Think about removing the last added item.

Explain what an underflow error is in the context of popping from a stack.

What happens if you try to take something from an empty box?

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