Pop Operation on Stack in DSA Python

Mental Model

A stack is like a pile of plates where you only take the top plate off. Pop means removing the top item.

Analogy: Imagine a stack of books on a table. You can only remove the top book first before taking any below it.

Top
 ↑
[3] -> [2] -> [1] -> null

Dry Run Walkthrough

Input: stack: 3 -> 2 -> 1 -> null, pop operation

Goal: Remove the top element (3) from the stack and show the new top

Step 1: Check if stack is empty

3 -> 2 -> 1 -> null

Why: We must ensure there is something to pop

Step 2: Remove the top element (3)

[pop->3] 2 -> 1 -> null

Why: Pop removes the top element and returns it

Step 3: Update top pointer to next element (2)

Top ↑ 2 -> 1 -> null

Why: The new top is now the element below the popped one

Result:

Top ↑ 2 -> 1 -> null
Popped element: 3

Annotated Code

DSA Python

class Node:
    def __init__(self, value):
        self.value = value
        self.next = None

class Stack:
    def __init__(self):
        self.top = None

    def push(self, value):
        new_node = Node(value)
        new_node.next = self.top
        self.top = new_node

    def pop(self):
        if self.top is None:
            return None  # stack empty
        popped_value = self.top.value
        self.top = self.top.next  # move top down
        return popped_value

    def __str__(self):
        curr = self.top
        values = []
        while curr:
            values.append(str(curr.value))
            curr = curr.next
        return ' -> '.join(values) + ' -> null'

# Driver code
stack = Stack()
stack.push(1)
stack.push(2)
stack.push(3)
print(stack)  # before pop
popped = stack.pop()
print(f"Popped element: {popped}")
print(stack)  # after pop

if self.top is None:

check if stack is empty before popping

popped_value = self.top.value

store the top element's value to return

self.top = self.top.next

update top pointer to next element, removing the current top

Output

3 -> 2 -> 1 -> null

Popped element: 3

2 -> 1 -> null

Complexity Analysis

Time: O(1) because pop removes the top element directly without traversal

Space: O(1) as no extra space is needed for pop operation

vs Alternative: Compared to removing from an array's start which is O(n), stack pop is faster and simpler

Edge Cases

empty stack

pop returns None and does not crash

DSA Python

if self.top is None:

stack with one element

pop removes the only element and top becomes None

DSA Python

self.top = self.top.next

When to Use This Pattern

When you see a problem requiring last-in-first-out removal, think of the stack pop operation to remove the top element efficiently.

Common Mistakes

Mistake: Trying to pop from an empty stack without checking causes errors
Fix: Always check if the stack is empty before popping

Mistake: Not updating the top pointer after pop leaves the stack in an incorrect state
Fix: Update the top pointer to the next node after popping

Summary

Removes the top element from the stack and returns it

Use when you need to remove the most recently added item first

The key is to update the top pointer to the next element after removal

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