Push Operation on Stack in DSA Python

Mental Model

A stack is like a pile where you add new items only on top. Pushing means placing a new item on the top of this pile.

Analogy: Imagine a stack of plates. When you add a new plate, you always put it on the top of the stack.

Top -> null

Dry Run Walkthrough

Input: Start with empty stack, push values 1, then 2, then 3

Goal: Add each new value on top of the stack so the last pushed is on top

Step 1: Push 1 onto empty stack

Top -> [1] -> null

Why: First item becomes the top of the stack

Step 2: Push 2 onto stack

Top -> [2] -> [1] -> null

Why: New item goes on top, previous top moves down

Step 3: Push 3 onto stack

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

Why: Keep adding new items on top, stack grows upwards

Result:

Top -> 3 -> 2 -> 1 -> null

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)  # create new node with value
        new_node.next = self.top  # new node points to current top
        self.top = new_node  # update top to new node

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

# Driver code
stack = Stack()
stack.push(1)
stack.push(2)
stack.push(3)
print(stack)

new_node = Node(value) # create new node with value

create a new node to hold the pushed value

new_node.next = self.top # new node points to current top

link new node to the current top node to keep stack chain

self.top = new_node # update top to new node

update stack top pointer to new node, making it the new top

Output

3 -> 2 -> 1 -> null

Complexity Analysis

Time: O(1) because push adds one item on top without traversing

Space: O(1) extra space for the new node created during push

vs Alternative: Compared to array-based stack, linked list push avoids resizing and copying, always O(1)

Edge Cases

Empty stack

Push sets top to new node correctly

DSA Python

self.top = new_node  # update top to new node

When to Use This Pattern

When you need to add an item to a stack quickly and always on top, use the push operation to maintain last-in-first-out order.

Common Mistakes

Mistake: Forgetting to link the new node's next to the current top
Fix: Always set new_node.next = self.top before updating self.top

Mistake: Updating top before linking new node's next
Fix: Link new_node.next first, then update self.top to new_node

Summary

Push adds a new item on top of the stack.

Use push when you want to insert elements in last-in-first-out order.

The key is to link the new node to the current top before updating the top pointer.