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

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after pushing elements on a stack?

Consider the following Python code that uses a list as a stack. What will be the printed output after all push operations?

DSA Python

stack = []
stack.append(5)
stack.append(10)
stack.append(15)
print(stack)

A [5, 15, 10]

B [15, 10, 5]

C [5, 10, 15]

D [10, 5, 15]

💡 Hint

Remember that append adds elements to the end of the list.

The append method adds elements to the end of the list, so the stack will be [5, 10, 15].

❓ Predict Output

intermediate

What is the top element after pushing on stack?

After pushing elements 3, 7, and 9 on an empty stack, what is the top element of the stack?

DSA Python

stack = []
stack.append(3)
stack.append(7)
stack.append(9)
top = stack[-1]
print(top)

A 9

B 7

C None

D 3

💡 Hint

The top element is the last one pushed.

The last element pushed is 9, so stack[-1] is 9.

🔧 Debug

advanced

Identify the error in this push operation code

What error will this code produce when trying to push an element onto the stack?

DSA Python

stack = []
stack.push(4)
print(stack)

A No error, output: [4]

B AttributeError: 'list' object has no attribute 'push'

C IndexError: list index out of range

D TypeError: push() missing 1 required positional argument

💡 Hint

Check the method name for adding elements to a list.

Python lists use append() to add elements, not push(). Using push() causes AttributeError.

❓ Predict Output

advanced

What is the stack content after multiple push operations?

Given this code, what is the final stack content printed?

DSA Python

stack = [1, 2]
stack.append(3)
stack.append(4)
stack.append(5)
print(stack)

A [3, 4, 5]

B [5, 4, 3, 2, 1]

C [1, 2, 5, 4, 3]

D [1, 2, 3, 4, 5]

💡 Hint

Appending adds elements to the end of the list.

Appending adds elements at the end, so the stack grows as [1, 2, 3, 4, 5].

🧠 Conceptual

expert

What is the time complexity of the push operation on a stack implemented with a Python list?

Choose the correct time complexity for the push operation (adding an element) on a stack implemented using a Python list with append().

A O(1) - Constant time

B O(n) - Linear time

C O(log n) - Logarithmic time

D O(n^2) - Quadratic time

💡 Hint

Appending to the end of a Python list is usually very fast.

Python list append() runs in average constant time O(1) because it adds element at the end.

Practice

1. What does the push operation do in a stack?

easy

A. Checks if the stack is empty

B. Removes the top element from the stack

C. Returns the bottom element of the stack

D. Adds an element to the top of the stack

Solution

1. Step 1: Understand stack behavior

   A stack follows Last In First Out (LIFO) principle, so new elements are added on top.

2. Step 2: Define push operation

   Push means adding a new element on the top of the stack.

3. Final Answer:

   Adds an element to the top of the stack -> Option D

4. Quick Check:

   Push = Add top element [OK]

Hint: Push always adds element on top of stack [OK]

Common Mistakes:

• Confusing push with pop operation
• Thinking push removes elements
• Mixing stack with queue operations

2. Which of the following is the correct syntax to push an element 5 into a stack named stack implemented as a Python list?

easy

A. stack.append(5)

B. stack.push(5)

C. stack.add(5)

D. stack.insert(0, 5)

Solution

1. Step 1: Recall Python list methods for stack

   Python lists use append() to add elements at the end, which acts as top of stack.

2. Step 2: Check each option

   push() and add() are not list methods; insert(0, 5) adds at bottom, not top.

3. Final Answer:

   stack.append(5) -> Option A

4. Quick Check:

   Use append() to push in Python list stack [OK]

Hint: Use append() to push in Python list stack [OK]

Common Mistakes:

• Using push() which is not a Python list method
• Using insert(0, value) which adds at bottom
• Confusing add() with append()

3. What will be the content of the stack after executing the following code?

stack = []
stack.append(10)
stack.append(20)
stack.append(30)

medium

A. [20, 10, 30]

B. [30, 20, 10]

C. [10, 20, 30]

D. [10, 30, 20]

Solution

1. Step 1: Trace each append operation

   Start with empty list. Append 10 -> [10], append 20 -> [10, 20], append 30 -> [10, 20, 30].

2. Step 2: Understand stack order

   Elements are added at the end; top is last element (30).

3. Final Answer:

   [10, 20, 30] -> Option C

4. Quick Check:

   Stack after pushes = [10, 20, 30] [OK]

Hint: Append adds at end; last element is top [OK]

Common Mistakes:

• Reversing the order of elements
• Confusing stack top with bottom
• Assuming append adds at start

4. Identify the error in the following code that tries to push an element 15 into a stack implemented as a list:

stack = [5, 10]
stack.push(15)

medium

A. Missing import statement for stack

B. Using push() method which does not exist for list

C. Stack should be initialized as a tuple

D. Using append() instead of push()

Solution

1. Step 1: Check list methods for adding elements

   Python lists do not have a push() method; they use append().

2. Step 2: Analyze the code error

   Calling stack.push(15) will cause an AttributeError because push is undefined.

3. Final Answer:

   Using push() method which does not exist for list -> Option B

4. Quick Check:

   List has append(), not push() [OK]

Hint: Use append() for list push, not push() [OK]

Common Mistakes:

• Assuming push() exists for Python lists
• Thinking stack must be tuple
• Ignoring AttributeError on push()

5. Given a stack implemented as a list, how would you push multiple elements [1, 2, 3] onto the stack so that 3 is on top after all pushes?

hard

A. Use a loop to append elements in order: 1, then 2, then 3

B. Use a loop to append elements in reverse order: 3, then 2, then 1

C. Use extend() method with reversed([1, 2, 3])

D. Insert elements at index 0 in order: 1, then 2, then 3

Solution

1. Step 1: Understand stack push order

   To have 3 on top, push 1 first, then 2, then 3 so 3 is last pushed.

2. Step 2: Analyze each option

   Loop append in order 1,2,3 puts 3 on top. Reverse order puts 1 on top. extend(reversed([1,2,3])) adds 3,2,1 so 1 on top. Insert at index 0 in order ends with [3,2,1], 1 on top.

3. Final Answer:

   Use a loop to append elements in order: 1, then 2, then 3 -> Option A

4. Quick Check:

   Push in order to get last element on top [OK]

Hint: Push elements in order to have last on top [OK]

Common Mistakes:

• Pushing in reverse order causing wrong top
• Using insert(0, val) which adds at bottom
• Using extend(reversed([1, 2, 3])) which puts wrong order on top