Push Operation on Stack in DSA Python - Execution Trace

Concept Flow - Push Operation on Stack

Create new element

↓

Check if stack is full?

Yes→Error: Stack Overflow

No↓

Place element on top

↓

Update top pointer

↓

Done

The push operation adds a new element on top of the stack if there is space, updating the top pointer.

Execution Sample

DSA Python

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

This code pushes 10 and then 20 onto the stack and prints the stack.

Execution Table

Step	Operation	Element Added	Stack State	Top Pointer Index
1	Initialize empty stack	None	[]	None
2	Push 10	10	[10]	0
3	Push 20	20	[10, 20]	1
4	End of operations	None	[10, 20]	1

💡 No more push operations; stack contains two elements with top at index 1.

Variable Tracker

Variable	Start	After Step 2	After Step 3	Final
stack	[]	[10]	[10, 20]	[10, 20]
top	None	0	1	1

Key Moments - 3 Insights

Why does the top pointer start at None and then become 0 after the first push?

Initially, the stack is empty so top has no valid index (None). After pushing the first element, top points to index 0 where the element is placed (see execution_table step 2).

What happens if we try to push when the stack is full?

The flow checks if the stack is full before pushing (concept_flow). If full, it stops with an error (stack overflow). This is not shown here because the stack is dynamic.

Why does the stack state show elements in order of insertion?

Because push adds elements on top, the last pushed element is at the end of the list (execution_table rows 2 and 3). The stack grows to the right.

Visual Quiz - 3 Questions

Look at the execution_table, what is the top pointer index after pushing 20?

A 0

B 1

C 2

D None

💡 Hint

Check the 'Top Pointer Index' column at step 3 in execution_table.

At which step does the stack first contain two elements?

A Step 1

B Step 2

C Step 3

D Step 4

💡 Hint

Look at the 'Stack State' column to see when two elements appear.

If we add a push operation for 30 after step 3, what will be the new top pointer index?

A 2

B 3

C 1

D 0

💡 Hint

Top pointer increases by 1 with each push (see variable_tracker for top).

Concept Snapshot

Push Operation on Stack:
- Adds new element on top
- Check if stack is full before push
- Place element at top + 1 index
- Update top pointer to new index
- Stack grows with last-in element on top

Full Transcript

The push operation on a stack adds a new element on top if there is space. We start with an empty stack and no top pointer. When we push the first element, it goes to index 0 and top points there. Each new push adds the element at the next index and updates top. If the stack is full, push cannot add more elements. The stack state shows elements in order of insertion, with the last pushed element on top. The top pointer tracks the index of the top element. This trace shows pushing 10 then 20, updating the stack and top pointer accordingly.

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