DSAverse

Basics & Data Structures

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Array Access

array[0..5]indexed access

Array

Stack

Queue

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Sorting

Trees

Graphs

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Beginner Friendly

Stack: Linked List Implementation

A dynamic stack using linked nodes. No fixed capacity — the top is always the head node.

Interactive Operations

Linked List Nodes

O(1) Operations

Stack Visualization

The HEAD node is always the TOP. Each node holds a value + a pointer to the next node.

Color Legend

DefaultActive / TargetCompletedError

Speed:

HEAD (TOP)

Stack is empty — push to begin

Size

NULL

Head

Dynamic

Allocation

Current Step

Use Push, Pop, or Peek to begin visualization.

Complexity Analysis

O(1)

Push

O(1)

Pop

O(1)

Peek

O(n) Space

Dynamic allocation + pointer per node

Linked List vs Array Stack

Dynamic Size

No fixed max — grows as needed

No Overflow

Limited only by available memory

Memory Efficient

Allocates exactly what is needed

Pointer Overhead

Extra memory per node for next reference

Cache Performance

Nodes scattered in memory (not contiguous)

How Operations Work

Push

Create node → set next = old head → update head to new node

Pop

Store head value → move head to head.next → return stored value

Peek

Return head.value without modifying the list

Test Yourself

Q1/3

In a linked list stack, where is the TOP of the stack?

Implementation

Python

class Node:
    def __init__(self, value):
        self.value = value
        self.next = None   # Reference to next node

class StackLinkedList:
    def __init__(self):
        self.head = None   # TOP of stack
        self.size = 0

    def push(self, value):
        new_node = Node(value)
        new_node.next = self.head  # Point to old top
        self.head = new_node       # New node is now top
        self.size += 1
        return value

    def pop(self):
        if self.head is None:
            raise Exception("Stack Underflow")
        value = self.head.value
        self.head = self.head.next  # Move top to next
        self.size -= 1
        return value

    def peek(self):
        if self.head is None:
            raise Exception("Stack is Empty")
        return self.head.value

    def is_empty(self):
        return self.head is None

    def get_size(self):
        return self.size

DSAverse

Basics & Data Structures

Loading Data Structures...

Array Access

array[0..5]indexed access

Array

Stack

Queue

Back to Basics

Beginner Friendly

Stack: Linked List Implementation

A dynamic stack using linked nodes. No fixed capacity — the top is always the head node.

Interactive Operations

Linked List Nodes

O(1) Operations

Stack Visualization

The HEAD node is always the TOP. Each node holds a value + a pointer to the next node.

Color Legend

DefaultActive / TargetCompletedError

Speed:

HEAD (TOP)

Stack is empty — push to begin

Size

NULL

Head

Dynamic

Allocation

Current Step

Use Push, Pop, or Peek to begin visualization.

Complexity Analysis

O(1)

Push

O(1)

Pop

O(1)

Peek

O(n) Space

Dynamic allocation + pointer per node

Linked List vs Array Stack

Dynamic Size

No fixed max — grows as needed

No Overflow

Limited only by available memory

Memory Efficient

Allocates exactly what is needed

Pointer Overhead

Extra memory per node for next reference

Cache Performance

Nodes scattered in memory (not contiguous)

How Operations Work

Push

Create node → set next = old head → update head to new node

Pop

Store head value → move head to head.next → return stored value

Peek

Return head.value without modifying the list

Test Yourself

Q1/3

In a linked list stack, where is the TOP of the stack?

Implementation

Python

class Node:
    def __init__(self, value):
        self.value = value
        self.next = None   # Reference to next node

class StackLinkedList:
    def __init__(self):
        self.head = None   # TOP of stack
        self.size = 0

    def push(self, value):
        new_node = Node(value)
        new_node.next = self.head  # Point to old top
        self.head = new_node       # New node is now top
        self.size += 1
        return value

    def pop(self):
        if self.head is None:
            raise Exception("Stack Underflow")
        value = self.head.value
        self.head = self.head.next  # Move top to next
        self.size -= 1
        return value

    def peek(self):
        if self.head is None:
            raise Exception("Stack is Empty")
        return self.head.value

    def is_empty(self):
        return self.head is None

    def get_size(self):
        return self.size