Pop Operation on Stack in DSA Python - Build from Scratch

Pop Operation on Stack

📖 Scenario: Imagine you have a stack of books. You can only take the top book off the stack at a time. This is how a stack data structure works in computers.

🎯 Goal: You will create a stack using a list, set up a variable to hold the stack size, perform a pop operation to remove the top item, and then print the stack after the pop.

📋 What You'll Learn

Create a list called stack with these exact elements in order: 10, 20, 30, 40, 50

Create a variable called stack_size that holds the length of the stack

Use the pop() method on stack to remove the top element

Print the stack after the pop operation

💡 Why This Matters

🌍 Real World

Stacks are used in many real-world applications like undo features in text editors, browser history, and managing function calls in programming.

💼 Career

Understanding stack operations is important for software developers, especially when working with algorithms, memory management, and system design.

1

Create the stack list

Create a list called stack with these exact elements in order: 10, 20, 30, 40, 50

DSA Python

# Create the stack list with the elements 10, 20, 30, 40, 50
# Your code here

Hint

Use square brackets to create a list and separate the numbers with commas.

2

Set the stack size variable

Create a variable called stack_size that holds the length of the stack list using the len() function

DSA Python

stack = [10, 20, 30, 40, 50]
# Create stack_size to hold the length of stack
# Your code here

Hint

Use len(stack) to get the number of items in the list.

3

Pop the top element from the stack

Use the pop() method on the stack list to remove the top element

DSA Python

stack = [10, 20, 30, 40, 50]
stack_size = len(stack)
# Remove the top element from stack using pop()
# Your code here

Hint

Call pop() on the list to remove the last item.

4

Print the stack after pop

Print the stack list after the pop operation to show its current state

DSA Python

stack = [10, 20, 30, 40, 50]
stack_size = len(stack)
stack.pop()
# Print the stack list
# Your code here

Hint

Use print(stack) to display the list.

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