Push Operation on Stack in DSA Python - Time & Space Complexity

Time Complexity: Push Operation on Stack

O(1)

Understanding Time Complexity

We want to understand how the time to add an item to a stack changes as the stack grows.

How does the push operation behave when the stack size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class Stack:
    def __init__(self):
        self.items = []

    def push(self, item):
        self.items.append(item)

This code adds an item to the top of the stack using a list's append method.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Adding one item to the end of the list (stack).
• How many times: Each push adds exactly one item, no loops or recursion involved.

How Execution Grows With Input

Adding one item takes the same amount of time no matter how many items are already in the stack.

Input Size (n)	Approx. Operations
10	1
100	1
1000	1

Pattern observation: The time stays constant as the stack grows.

Final Time Complexity

Time Complexity: O(1)

This means adding an item to the stack takes the same short time no matter how big the stack is.

Common Mistake

[X] Wrong: "Push takes longer as the stack grows because it has to move all items."

[OK] Correct: The push just adds to the end of the list without moving existing items, so time stays the same.

Interview Connect

Knowing push is very fast helps you explain why stacks are great for quick add/remove tasks in real programs.

Self-Check

"What if the stack was implemented using a linked list instead of a list? How would the time complexity of push change?"

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