🧠 Introduction: Mastering Linked Lists From Scratch, Like a Human

Stepping into the world of Data Structures? You’ve likely encountered the term ‘Linked List’ and might feel it’s a concept reserved for seasoned programmers. But let’s demystify it: a linked list is simply a series of interconnected ‘nodes,’ each holding a piece of data and a pointer to the subsequent node. Imagine a treasure hunt where each clue leads you to the next, until you find the final prize. That’s a linked list! Unlike arrays, which store elements in contiguous memory blocks, linked lists are dynamically allocated, allowing nodes to reside anywhere in memory and connect through references. This flexibility makes them invaluable for scenarios where data size frequently changes. In this guide, we’ll explore linked lists from the ground up, providing intuitive explanations, visual analogies, and practical C++ examples. Our goal isn’t just to help you write linked list code, but to foster a deep understanding of their underlying mechanics, making you a confident and articulate developer.

What is a Linked List?

At its core, a Linked List is a fundamental linear data structure. Think of it as a chain where each link is an ‘element’ or ‘node.’ These nodes aren’t stored side-by-side in memory like array elements; instead, they are linked together through references, often called pointers. Every node in a singly linked list typically consists of two main parts:

• Data: This is where the actual value or information is stored.
• Next (Pointer): This acts as an address, pointing to the next node in the sequence.

This structure offers significant advantages, primarily its dynamic nature, meaning memory is allocated as needed during program execution, providing great flexibility compared to static arrays.

Structure of a Node

Before diving into operations, let’s understand the building block of any linked list: the Node. Each node is a self-contained unit that holds both its own data and a reference to the next node in the sequence. In C++, this can be represented as a class:

// Definition of a Node class used in a Linked List
class Node {
public:
    int data;      // Stores the value or data part of the node
    Node* next;    // Pointer to the next node in the linked list

    // Constructor to initialize a new node with a given value
    Node(int val) {
        data = val;   // Assign the passed value to the data field
        next = NULL;  // Initially, the next pointer is set to NULL (no next node yet)
    }
};

Here, data stores an integer value (though it could be any data type), and next is a pointer that holds the memory address of the subsequent node. If next is NULL, it signifies the end of the list.

Basic Operations on Singly Linked List

1️⃣ Create and Print a Linked List

📘 Concept:

Dynamically allocate n nodes and establish connections between them using next pointers.

🔍 How it Works

Imagine you’re building a train. Each compartment (node) holds passengers (data) and has a door (pointer) to the next compartment.

1. You create the very first compartment; this is your head (the starting point of your train).
2. As you add new compartments, you connect each new one to the door of the previous compartment.

To print the list, you simply start from the head and ‘walk’ through each node by following its next pointer, displaying the data at each stop until you reach a NULL pointer (the end of the train).

💡 Real-World Analogy

It’s like following a trail of breadcrumbs; each crumb (node) leads you to the next, revealing the path (list) until the trail ends.

💻 Code:

#include <iostream>
using namespace std;

// Definition of the Node class (represents a single element in the linked list)
class Node {
public:
    int data;       // Stores the data/value of the node
    Node* next;     // Pointer to the next node in the linked list

    // Constructor to initialize a node with a given value
    Node(int val) {
        data = val;   // Assign the input value to the data part
        next = NULL;  // Initially, there is no next node (so next is NULL)
    }
};

// Function to print all elements of the linked list
void printList(Node* head) {
    Node* temp = head;  // Temporary pointer to traverse the list
    while (temp != NULL) {  // Loop until we reach the end of the list
        cout << temp->data << " ";  // Print the data of the current node
        temp = temp->next;          // Move to the next node
    }
    cout << endl;  // Print a newline after the entire list is displayed
}

int main() {
    Node* head = NULL;  // Pointer to the first node (head) of the linked list
    Node* tail = NULL;  // Pointer to the last node (tail) to easily add new nodes
    int n;              // Number of nodes the user wants to create

    cout << "Enter number of nodes: ";
    cin >> n;  // Input how many nodes to create

    // Loop to create 'n' nodes
    for (int i = 0; i < n; i++) {
        int val;
        cout << "Enter value: ";
        cin >> val;  // Input the data value for each node

        Node* newNode = new Node(val);  // Dynamically create a new node

        // If the list is empty, the new node becomes both head and tail
        if (head == NULL) {
            head = tail = newNode;
        } 
        // Otherwise, link the new node to the end of the list
        else {
            tail->next = newNode;  // Connect the last node to the new one
            tail = newNode;        // Update tail to point to the new last node
        }
    }
    cout << "Linked List: ";
    printList(head);

    return 0; 
}

🧪 Example:

Input:
Enter number of nodes: 3
Enter value: 10
Enter value: 20
Enter value: 30
Output:
Linked List: 10 20 30

2️⃣ Count Nodes

📘 Problem:

Determine the total number of nodes in a given linked list.

🔍 Logic:

This is a straightforward traversal.

1. Initialize a counter to zero.
2. Start from the head of the list.
3. As you visit each node, increment the counter.
4. Continue until you reach the end of the list (i.e., your current node pointer is NULL).
5. The final value of the counter will be your node count.

Analogy: Counting the number of passengers in a train by walking through each compartment.

💻 Code:

// Function to count the total number of nodes in a linked list
int countNodes(Node* head) {
    int count = 0;        // Variable to keep track of the number of nodes
    Node* temp = head;    // Temporary pointer to traverse the list (starting from head)

    // Traverse the entire linked list until we reach the end (NULL)
    while (temp != NULL) {
        count++;          // Increment count for each node visited
        temp = temp->next; // Move to the next node
    }

    // Return the total count after traversal
    return count;
}

3️⃣ Insert a Node at the Beginning

🔍 How it Works

You’re essentially placing a new ‘first car’ on your train.

💡 Logic:

1. Create your newNode with the data you want to insert.
2. Make this newNode‘s next pointer point to the current head of the list. This links your new node to the rest of the chain.
3. Update the head of the list to now point to your newNode, as it’s the new beginning.

💻 Code:

// Function to insert a new node at the beginning of the linked list
void insertAtBeginning(Node* &head, int value) {
    // Step 1: Create a new node with the given value
    Node* newNode = new Node(value);

    // Step 2: Point the new node's 'next' to the current head of the list
    // (This makes the new node link to the previous first node)
    newNode->next = head;

    // Step 3: Move the head pointer to point to the new node
    // (Now the new node becomes the first node in the list)
    head = newNode;
}

4️⃣ Insert a Node at the End

🔍 How it Works

You’re adding a new compartment to the very end of your existing train.

💡 Logic:

1. Create your newNode.
2. Special Case: If the list is empty (head == NULL), the newNode simply becomes the head.
3. Otherwise, you must traverse the list, starting from the head, until you find the last node. The last node is identified by its next pointer being NULL.
4. Once at the last node, update its next pointer to point to your newNode.
5. The newNode‘s next pointer should remain NULL (it’s the new end).

💻 Code:

// Function to insert a new node at the end of the linked list
void insertAtEnd(Node* &head, int value) {
    // Step 1: Create a new node with the given value
    Node* newNode = new Node(value);

    // Step 2: If the list is empty, make the new node the head
    if (head == NULL) {
        head = newNode;  // The new node becomes the first and only node
        return;          // Exit the function since insertion is done
    }

    // Step 3: Otherwise, traverse to the last node
    Node* temp = head;
    while (temp->next != NULL)  // Move forward until we find the last node
        temp = temp->next;

    // Step 4: Link the last node to the new node
    temp->next = newNode;
}

5️⃣ Insert at Specific Position

🔍 How it Works

This is like inserting a new car into the middle of a train. You need to detach two cars, insert the new one, and then re-attach the second car to the new one.

💡 Logic:

1. Edge Case: If pos is 1, simply use the insertAtBeginning function.
2. Create your newNode.
3. Traverse the list using a temp pointer until it reaches the node before the desired insertion pos-1.
4. Now, the newNode needs to point to the node that temp was originally pointing to (i.e., newNode->next = temp->next).
5. Then, temp needs to point to the newNode (i.e., temp->next = newNode).

// Function to insert a new node at a specific position in the linked list
void insertAtPosition(Node* &head, int value, int pos) {
    // Step 1: If position is 1, insert the new node at the beginning
    if (pos == 1) {
        insertAtBeginning(head, value);  // Reuse the function you already created
        return;  // Exit after insertion
    }

    // Step 2: Create a new node with the given value
    Node* newNode = new Node(value);

    // Step 3: Traverse the list to find the (pos-1)th node
    Node* temp = head;
    for (int i = 1; i < pos - 1 && temp != NULL; i++)
        temp = temp->next;

    // Step 4: If temp becomes NULL, it means the position is invalid
    if (temp == NULL)
        return; // Position out of bounds

    // Step 5: Adjust the links to insert the new node at the desired position
    newNode->next = temp->next;  // Link new node to the next node
    temp->next = newNode;        // Link previous node to the new node
}

6️⃣ Delete First Node

📘 Problem:

Eliminate the head node from the linked list.

🔍 Logic:

1. Edge Case: If the list is empty (head == NULL), there’s nothing to delete.
2. Create a temporary pointer, say temp, and make it point to the current head. This is so we can delete the old head later and free its memory.
3. Advance the head pointer to the next node in the list (head = head->next). Now, what was the second node is the new first node.
4. Finally, use delete temp to deallocate the memory of the original head node.

Analogy: Detaching the first car from a train. The second car then becomes the new first car.

// Function to delete the first node (head) of the linked list
void deleteFirst(Node* &head) {
    // Step 1: Check if the list is empty
    if (head == NULL)
        return;  // Nothing to delete if the list is already empty

    // Step 2: Store the current head node in a temporary pointer
    Node* temp = head;

    // Step 3: Move the head pointer to the next node
    head = head->next;

    // Step 4: Delete the old head node to free memory
    delete temp;
}

7️⃣ Delete Last Node

📘 Problem:

Remove the last node from the linked list.

🔍 Logic:

1. Edge Case 1: If the list is empty (head == NULL), do nothing.
2. Edge Case 2: If there’s only one node (head->next == NULL), delete the head and set head to NULL.
3. Otherwise, traverse the list with a temp pointer until temp->next->next == NULL. This means temp is now pointing to the second-to-last node.
4. The node to be deleted is temp->next. Store this in a temporary pointer (e.g., toDelete).
5. Set temp->next = NULL to break the link to the last node.
6. Finally, delete toDelete to free the memory of the former last node.

Analogy: Removing the last compartment from a train. You need to be in the second-to-last compartment to disconnect it.

// Function to delete the last node of the linked list
void deleteLast(Node* &head) {
    // Step 1: If the list is empty, nothing to delete
    if (head == NULL)
        return;

    // Step 2: If there is only one node, delete it and make head NULL
    if (head->next == NULL) {
        delete head;   // Free memory of the single node
        head = NULL;   // Set head to NULL (list becomes empty)
        return;
    }

    // Step 3: Traverse the list to find the second last node
    Node* temp = head;
    while (temp->next->next != NULL)  // Stop one node before the last
        temp = temp->next;

    // Step 4: Delete the last node and set the second last node’s next to NULL
    delete temp->next;   // Free memory of the last node
    temp->next = NULL;   // Mark the new end of the list
}

8️⃣ Find Maximum and Minimum Value

📘 Problem:

Identify and display the smallest and largest data values present in the linked list.

🔍 Logic:

1. Edge Case: If the list is empty, there are no values to compare.
2. Initialize minVal and maxVal with the data of the head node. This assumes at least one node.
3. Traverse the list starting from the head.
4. For each node visited, compare its data with minVal and maxVal:
   • If node->data is less than minVal, update minVal.
   • If node->data is greater than maxVal, update maxVal.
5. After the traversal completes, minVal and maxVal will hold the true minimum and maximum values.

Analogy: Scanning through a list of numbers to find the highest and lowest scores.

// Function to find and print the minimum and maximum values in a linked list
void findMinMax(Node* head) {
    // Step 1: Check if the list is empty
    if (head == NULL) {
        cout << "List is empty, no min/max." << endl; // Added a message for empty list
        return;  // No nodes to process
    }

    // Step 2: Initialize min and max with the first node’s data
    int minVal = head->data;
    int maxVal = head->data;

    // Step 3: Create a temporary pointer to traverse the list
    Node* temp = head;

    // Step 4: Traverse the entire linked list
    while (temp != NULL) {
        // Update min and max values if current node’s data is smaller/larger
        minVal = min(minVal, temp->data);
        maxVal = max(maxVal, temp->data);

        // Move to the next node
        temp = temp->next;
    }

    // Step 5: Print the final results
    cout << "Min = " << minVal << ", Max = " << maxVal << endl;
}

9️⃣ Find Middle Node (Fast & Slow Pointer)

📘 Problem:

Identify the middle node of a linked list in a single traversal without first counting the total nodes.

🔍 Logic:

This approach uses two pointers moving at different speeds:

1. Initialize two pointers, slow and fast, both starting at the head of the list.
2. In a loop, advance slow by one step (slow = slow->next) and fast by two steps (fast = fast->next->next) in each iteration.
3. The loop continues as long as fast and fast->next are not NULL (to handle both even and odd length lists).
4. When the fast pointer reaches the end of the list (or NULL), the slow pointer will naturally be positioned at the middle node.

Analogy: Imagine two runners on a track. One runs twice as fast as the other. When the faster runner finishes a full lap, the slower runner will be exactly halfway.

// Function to find and print the middle node of a linked list
void findMiddle(Node* head) {
    // Step 1: Handle empty or single-node list
    if (head == NULL) {
        cout << "List is empty, no middle node." << endl; // Added message for empty list
        return;
    }
    if (head->next == NULL) { // Single node list
        cout << "Middle Node: " << head->data << endl;
        return;
    }

    // Step 2: Initialize two pointers - slow and fast
    Node* slow = head;  // Moves one step at a time
    Node* fast = head;  // Moves two steps at a time

    // Step 3: Traverse the list
    // When 'fast' reaches the end, 'slow' will be at the middle
    while (fast != NULL && fast->next != NULL) {
        slow = slow->next;         // Move slow pointer by one node
        fast = fast->next->next;   // Move fast pointer by two nodes
    }

    // Step 4: Print the middle node's data
    cout << "Middle Node: " << slow->data << endl;
}

🔟 Reverse a Linked List (Iterative)

📘 Problem:

Invert the order of nodes in a singly linked list.

🔍 Logic:

This operation requires three pointers to keep track of the current node, the previous node, and the next node in the original sequence.

1. Initialize prev to NULL (the new tail will point to nothing).
2. Initialize curr to head (start from the current beginning).
3. Initialize next to NULL (a temporary placeholder).
4. Iterate through the list while curr is not NULL:
   • Store curr->next in next to save the link to the rest of the list.
   • Change curr->next to prev, reversing the pointer direction.
   • Move prev to curr (the current node becomes the ‘previous’ for the next iteration).
   • Move curr to next (advance to the next node in the original list).
5. After the loop, curr will be NULL, and prev will be pointing to the last node of the original list (which is now the new head). Update head = prev.

Analogy: Flipping a train around. Each car’s connection to the next needs to be re-wired to point backward.

// Function to reverse a linked list
void reverseList(Node* &head) {
    // Step 1: Initialize three pointers
    Node* prev = NULL;   // Points to the previous node (initially none)
    Node* curr = head;   // Points to the current node (starts at head)
    Node* next = NULL;   // Temporarily stores the next node

    // Step 2: Traverse the list and reverse the links
    while (curr != NULL) {
        next = curr->next;   // Store the next node
        curr->next = prev;   // Reverse the link
        prev = curr;         // Move prev one step forward
        curr = next;         // Move curr one step forward
    }

    // Step 3: Update the head to point to the new first node
    head = prev;
}

🧩 Complete Linked List Program (with All Functions + Example Outputs)

To consolidate our learning, here is a complete C++ program incorporating all the basic singly linked list operations we’ve discussed, along with an example main function to demonstrate their usage and outputs. This provides a hands-on way to see how these functions interact and transform the linked list.

#include <iostream>
#include <algorithm>  // for min() and max()
using namespace std;

// -------------------- NODE STRUCTURE --------------------
class Node {
public:
    int data;      // To store data of the node
    Node* next;    // Pointer to next node

    // Constructor to initialize node with a value
    Node(int val) {
        data = val;
        next = NULL;
    }
};

// -------------------- PRINT LINKED LIST --------------------
void printList(Node* head) {
    Node* temp = head;
    while (temp != NULL) {
        cout << temp->data << " ";
        temp = temp->next;
    }
    cout << endl;
}

// -------------------- COUNT NODES --------------------
int countNodes(Node* head) {
    int count = 0;
    Node* temp = head;
    while (temp != NULL) {
        count++;
        temp = temp->next;
    }
    return count;
}

// -------------------- INSERT AT BEGINNING --------------------
void insertAtBeginning(Node* &head, int value) {
    Node* newNode = new Node(value);
    newNode->next = head;
    head = newNode;
}

// -------------------- INSERT AT END --------------------
void insertAtEnd(Node* &head, int value) {
    Node* newNode = new Node(value);
    if (head == NULL) { head = newNode; return; }

    Node* temp = head;
    while (temp->next != NULL)
        temp = temp->next;
    temp->next = newNode;
}

// -------------------- INSERT AT POSITION --------------------
void insertAtPosition(Node* &head, int value, int pos) {
    if (pos == 1) {
        insertAtBeginning(head, value);
        return;
    }
    Node* newNode = new Node(value);
    Node* temp = head;
    for (int i = 1; i < pos - 1 && temp != NULL; i++)
        temp = temp->next;
    if (temp == NULL) return;
    newNode->next = temp->next;
    temp->next = newNode;
}

// -------------------- DELETE FIRST NODE --------------------
void deleteFirst(Node* &head) {
    if (head == NULL) return;
    Node* temp = head;
    head = head->next;
    delete temp;
}

// -------------------- DELETE LAST NODE --------------------
void deleteLast(Node* &head) {
    if (head == NULL) return;
    if (head->next == NULL) { delete head; head = NULL; return; }

    Node* temp = head;
    while (temp->next->next != NULL)
        temp = temp->next;
    delete temp->next;
    temp->next = NULL;
}

// -------------------- FIND MIN & MAX --------------------
void findMinMax(Node* head) {
    if (head == NULL) { cout << "List is empty, no min/max." << endl; return; }
    int minVal = head->data, maxVal = head->data; 
    Node* temp = head;
    while (temp != NULL) {
        minVal = min(minVal, temp->data);
        maxVal = max(maxVal, temp->data);
        temp = temp->next;
    }
    cout << "Min = " << minVal << ", Max = " << maxVal << endl;
}

// -------------------- FIND MIDDLE NODE --------------------
void findMiddle(Node* head) {
    if (head == NULL) { cout << "List is empty, no middle node." << endl; return; }
    if (head->next == NULL) { cout << "Middle Node: " << head->data << endl; return; }
    Node* slow = head;
    Node* fast = head;
    while (fast != NULL && fast->next != NULL) {
        slow = slow->next;
        fast = fast->next->next;
    }
    cout << "Middle Node: " << slow->data << endl;
}

// -------------------- REVERSE LINKED LIST --------------------
void reverseList(Node* &head) {
    Node* prev = NULL;
    Node* curr = head;
    Node* next = NULL;
    while (curr != NULL) {
        next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    head = prev;
}

// -------------------- MAIN FUNCTION --------------------
int main() {
    Node* head = NULL;

    cout << "Creating Linked List: 10 -> 20 -> 30 -> 40 -> 50
";
    insertAtEnd(head, 10);
    insertAtEnd(head, 20);
    insertAtEnd(head, 30);
    insertAtEnd(head, 40);
    insertAtEnd(head, 50);

    cout << "Initial List: ";
    printList(head);

    cout << "
[1] Insert at Beginning (5): ";
    insertAtBeginning(head, 5);
    printList(head);

    cout << "
[2] Insert at End (60): ";
    insertAtEnd(head, 60);
    printList(head);

    cout << "
[3] Insert at Position (25 at pos 4): ";
    insertAtPosition(head, 25, 4);
    printList(head);

    cout << "
[4] Delete First Node: ";
    deleteFirst(head);
    printList(head);

    cout << "
[5] Delete Last Node: ";
    deleteLast(head);
    printList(head);

    cout << "
[6] Count of Nodes: " << countNodes(head) << endl;

    cout << "
[7] Find Min & Max: ";
    findMinMax(head);

    cout << "
[8] Find Middle: ";
    findMiddle(head);

    cout << "
[9] Reverse Linked List: ";
    reverseList(head);
    printList(head);

    // Clean up memory to prevent leaks
    Node* current = head;
    while (current != NULL) {
        Node* nextNode = current->next;
        delete current;
        current = nextNode;
    }

    return 0;
}

🧪 Example Output:

Creating Linked List: 10 -> 20 -> 30 -> 40 -> 50
Initial List: 10 20 30 40 50 

[1] Insert at Beginning (5): 5 10 20 30 40 50 

[2] Insert at End (60): 5 10 20 30 40 50 60 

[3] Insert at Position (25 at pos 4): 5 10 20 25 30 40 50 60 

[4] Delete First Node: 10 20 25 30 40 50 60 

[5] Delete Last Node: 10 20 25 30 40 50 

[6] Count of Nodes: 6

[7] Find Min & Max: Min = 10, Max = 50

[8] Find Middle: Middle Node: 30

[9] Reverse Linked List: 50 40 30 25 20 10 

🧠 Key Takeaways

After exploring the fundamental operations, let’s distill the most crucial concepts for truly mastering linked lists:

• 1. Linked Lists are Dynamic & Connected: The defining feature of linked lists is their flexibility. Nodes are not fixed in memory; they are scattered and linked by pointers, making them ideal for data whose size changes frequently.
• 2. The Head is Your Anchor: The head pointer is the gateway to your entire list. Always protect and correctly update it during insertions or deletions, as losing the head means losing access to your data.
• 3. Pointers are Everything: Every operation on a linked list boils down to manipulating next pointers. Visualize these connections—how they form, break, and redirect—to understand the logic.
• 4. Visual Learning is Key: Don’t just memorize code. Draw out your linked list on paper, trace pointer movements step-by-step, and actively visualize memory changes. This deepens comprehension far more than rote memorization.
• 5. Master Edge Cases: Always consider empty lists, single-node lists, and operations at the beginning or end. Robust linked list code handles these scenarios gracefully.
• 6. The Power of Two Pointers: Techniques like the “fast and slow pointer” are incredibly powerful for solving various linked list problems efficiently (e.g., finding the middle, detecting cycles).
• 7. Memory Management (C++ Specific): In C++, remember to pair every new with a `delete` to prevent memory leaks. Properly deallocate nodes that are no longer needed.

💡 Things to Keep in Mind

Here are some practical tips to ensure your linked list implementations are robust and error-free:

• ⚙️ 1. Initialize Pointers Carefully: Always set new pointers (especially next for a newly created node, or prev in reversal algorithms) to NULL initially. This prevents unexpected behavior.
• 🧩 2. Don’t Lose Your Way: When modifying links, especially during deletion or reversal, ensure you temporarily store the necessary next pointer before overwriting it. Otherwise, you might lose access to the rest of your list.
• 🔄 3. Safe Traversal is Crucial: Always include checks like temp == NULL before accessing temp->data or temp->next to avoid dereferencing NULL pointers, which leads to crashes.
• 🧹 4. Debug with Visualization: If your code isn’t working as expected, mentally (or physically) draw the list. Step through your code line by line, updating your drawing, to pinpoint where pointers are going wrong.
• 🎯 5. Understand Before You Code: Before writing a single line of code for an operation, clearly define the steps involved in manipulating the pointers. A solid conceptual understanding simplifies coding significantly.

🧩 Singly Linked List Practice Questions

To truly solidify your understanding, practice is indispensable. Here’s a comprehensive set of singly linked list practice questions, categorized by topic:

🔹 Part 1: Basics

1. Create a singly linked list with n nodes, take input for each node, and print the list.
2. Count the number of nodes in a linked list.
3. Find the sum of all node values in a linked list.
4. Display the linked list in reverse order (without actually reversing it).

🔹 Part 2: Insertion

1. Insert a node at the beginning of a linked list.
2. Insert a node at the end of a linked list.
3. Insert a node at a specific position in the linked list.
4. Insert a node after a given value in the linked list.

🔹 Part 3: Deletion

1. Delete the first node of a linked list.
2. Delete the last node of a linked list.
3. Delete a node at a specific position.
4. Delete a node by its value.

🔹 Part 4: Searching & Traversal

1. Search for a given value in the linked list and return its position.
2. Find the maximum and minimum value in the linked list.
3. Find the middle element of a linked list using the fast and slow pointer approach.
4. Count how many times a given value occurs in the linked list.

🔹 Part 5: Reversal & Modifications

1. Reverse a linked list iteratively.
2. Reverse a linked list recursively.
3. Remove duplicate elements from a sorted linked list.
4. Remove duplicate elements from an unsorted linked list.

🔹 Part 6: Special Operations

1. Merge two sorted linked lists into one sorted list.
2. Find the Nth node from the end of a linked list.
3. Check whether a linked list is palindromic (reads the same forward and backward).
4. Detect a cycle/loop in a linked list using Floyd’s Cycle Detection Algorithm.
5. Remove a cycle from a linked list if it exists.

🔹 Part 7: Advanced Practice

1. Split a linked list into two halves.
2. Find the intersection point of two linked lists.
3. Sort a linked list using merge sort.
4. Add two numbers represented by linked lists.
5. Clone a linked list with random pointers (advanced concept).

✨ Wrapping Up

Congratulations! You’ve successfully laid the groundwork for one of the most fundamental data structures in computer science: the Singly Linked List. You now understand its structure, basic operations, and key principles. While pointers might feel intricate at first, consistent practice and visualization will transform them into intuitive tools in your programming arsenal.

This is just the beginning of your journey. Next, you can delve into Doubly Linked Lists, which offer enhanced flexibility by allowing traversal in both directions, simplifying certain operations even further.

Keep practicing, keep visualizing, and get ready to expand your data structure knowledge!

✍️ About the Author

Hey there! I’m Manas, a Computer Science student passionate about breaking down complex CS topics into simple, human explanations.

I love teaching data structures, working on tech projects, and helping beginners learn to think like programmers.

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