Insert at Beginning Head Insert in DSA Java - Build from Scratch

Insert at Beginning Head Insert

📖 Scenario: Imagine you are managing a simple list of tasks to do. You want to add new tasks at the very start of your list so that the newest task is always done first.

🎯 Goal: You will build a simple linked list and learn how to insert a new node at the beginning (head) of the list.

📋 What You'll Learn

Create a Node class with an integer data and a next pointer

Create a LinkedList class with a head pointer

Write a method to insert a new node at the beginning of the list

Write a method to print the list from head to end

💡 Why This Matters

🌍 Real World

Linked lists are used in many applications like task scheduling, undo functionality in editors, and managing playlists where order and quick insertions matter.

💼 Career

Understanding linked lists and how to insert nodes at the beginning is a fundamental skill for software developers, especially for roles involving data structure optimization and system design.

1

Create Node and LinkedList classes

Create a class called Node with an integer variable data and a Node variable next. Then create a class called LinkedList with a Node variable head initialized to null.

DSA Java

// Your code here

Hint

Think of Node as a box holding a number and a pointer to the next box. The LinkedList keeps track of the first box with head.

2

Add insertAtBeginning method

Inside the LinkedList class, add a method called insertAtBeginning that takes an integer data as input. This method should create a new Node with the given data and insert it at the start of the list by updating head.

DSA Java

class Node {
    int data;
    Node next;

    Node(int data) {
        this.data = data;
        this.next = null;
    }
}

class LinkedList {
    Node head = null;

    // Your code here
}

Hint

To insert at the beginning, point the new node's next to the current head, then update head to the new node.

3

Add printList method

Inside the LinkedList class, add a method called printList that prints all node data from head to the end, separated by arrows ->, ending with null.

DSA Java

class Node {
    int data;
    Node next;

    Node(int data) {
        this.data = data;
        this.next = null;
    }
}

class LinkedList {
    Node head = null;

    void insertAtBeginning(int data) {
        Node newNode = new Node(data);
        newNode.next = head;
        head = newNode;
    }

    // Your code here
}

Hint

Use a temporary pointer to walk through the list from head and print each node's data followed by an arrow.

4

Test insertion and print the list

In the main method, create a LinkedList object called list. Insert the integers 10, 20, and 30 at the beginning using insertAtBeginning. Then call printList to display the list.

DSA Java

class Node {
    int data;
    Node next;

    Node(int data) {
        this.data = data;
        this.next = null;
    }
}

class LinkedList {
    Node head = null;

    void insertAtBeginning(int data) {
        Node newNode = new Node(data);
        newNode.next = head;
        head = newNode;
    }

    void printList() {
        Node current = head;
        while (current != null) {
            System.out.print(current.data + " -> ");
            current = current.next;
        }
        System.out.print("null\n");
    }
}

public class Main {
    public static void main(String[] args) {
        // Your code here
    }
}

Hint

Remember, inserting at the beginning means the last inserted number appears first when printing.

Practice

1. What does the "Insert at Beginning" operation do in a singly linked list?

easy

A. Deletes the last node of the list

B. Adds a new node at the end of the list

C. Adds a new node at the start of the list

D. Deletes the first node of the list

Solution

1. Step 1: Understand the operation

   Inserting at the beginning means placing a new node before the current first node.

2. Step 2: Effect on the list

   The new node becomes the head, pushing the old head to second position.

3. Final Answer:

   Adds a new node at the start of the list -> Option C

4. Quick Check:

   Insert at beginning = Add new head [OK]

Hint: New node becomes head, old head moves next [OK]

Common Mistakes:

• Confusing insert at beginning with insert at end
• Thinking it deletes nodes
• Assuming it changes the tail node

2. Which of the following is the correct Java syntax to insert a new node at the beginning of a singly linked list named list with value 5?

easy

A. list.insertEnd(5);

B. list.add(5);

C. list.head.next = new Node(5);

D. list.head = new Node(5, list.head);

Solution

1. Step 1: Understand Node constructor

   Assuming Node constructor takes value and next node, new Node(5, list.head) creates a new node pointing to current head.

2. Step 2: Assign new node as head

   Setting list.head = new Node(5, list.head); inserts at beginning correctly.

3. Final Answer:

   list.head = new Node(5, list.head); -> Option D

4. Quick Check:

   New node points to old head, becomes new head [OK]

Hint: New node points to old head, assign new node as head [OK]

Common Mistakes:

• Using add() which may add at end
• Changing next of head instead of head itself
• Using insertEnd() which adds at tail

3. What will be the output linked list after executing the following Java code?

class Node {
  int data;
  Node next;
  Node(int d, Node n) { data = d; next = n; }
}

Node head = new Node(3, null);
head = new Node(2, head);
head = new Node(1, head);

// Print list
Node temp = head;
while(temp != null) {
  System.out.print(temp.data + " -> ");
  temp = temp.next;
}
System.out.print("null");

medium

A. 3 -> 2 -> 1 -> null

B. 1 -> 2 -> 3 -> null

C. null -> 1 -> 2 -> 3

D. 2 -> 1 -> 3 -> null

Solution

1. Step 1: Trace insertions

   First node: 3 -> null
   Insert 2 at beginning: 2 -> 3 -> null
   Insert 1 at beginning: 1 -> 2 -> 3 -> null

2. Step 2: Print list

   Loop prints nodes in order: 1, 2, 3, then null.

3. Final Answer:

   1 -> 2 -> 3 -> null -> Option B

4. Quick Check:

   Head insertions reverse order: 1,2,3 [OK]

Hint: Head insert reverses order of inserted nodes [OK]

Common Mistakes:

• Assuming insertion adds at end
• Printing nodes in wrong order
• Confusing next pointers

4. Identify the error in the following code snippet that tries to insert a new node with value 10 at the beginning of a singly linked list.

class Node {
  int data;
  Node next;
  Node(int d) { data = d; next = null; }
}

Node head = null;
Node newNode = new Node(10);
newNode.next = head.next;
head = newNode;

medium

A. Accessing head.next when head is null causes NullPointerException

B. New node is not assigned to head

C. Node constructor is missing next parameter

D. No error, code works fine

Solution

1. Step 1: Analyze head initialization

   Head is null initially, so head.next access is invalid and causes error.

2. Step 2: Identify cause of error

   Trying to assign newNode.next = head.next; throws NullPointerException because head is null.

3. Final Answer:

   Accessing head.next when head is null causes NullPointerException -> Option A

4. Quick Check:

   Null head means head.next invalid [OK]

Hint: Check if head is null before accessing head.next [OK]

Common Mistakes:

• Ignoring null head before accessing next
• Assuming constructor error
• Thinking newNode not assigned to head

5. You have a singly linked list with nodes containing values [4, 5, 6] in that order. You want to insert a new node with value 3 at the beginning. After insertion, you want to print the list in the format shown below.

Which code snippet correctly performs the insertion and prints the list?

class Node {
  int data;
  Node next;
  Node(int d) { data = d; next = null; }
}

Node head = new Node(4);
head.next = new Node(5);
head.next.next = new Node(6);

// Insert new node with value 3 at beginning and print list
// Fill in the blanks below

// Insert code here

// Print code here

hard

A. Node newNode = new Node(3);
newNode.next = head;
head = newNode;
Node temp = head;
while(temp != null) {
System.out.print(temp.data + " -> ");
temp = temp.next;
}
System.out.print("null");

B. Node newNode = new Node(3);
head.next = newNode;
newNode.next = null;
Node temp = head;
while(temp != null) {
System.out.print(temp.data + " -> ");
temp = temp.next;
}
System.out.print("null");

C. Node newNode = new Node(3);
head = newNode;
newNode.next = head.next;
Node temp = head;
while(temp != null) {
System.out.print(temp.data + " -> ");
temp = temp.next;
}
System.out.print("null");

D. Node newNode = new Node(3);
newNode.next = null;
head.next = newNode;
Node temp = head;
while(temp != null) {
System.out.print(temp.data + " -> ");
temp = temp.next;
}
System.out.print("null");

Solution

1. Step 1: Insert new node at beginning

   Create new node with value 3, set its next to current head, then update head to new node.

2. Step 2: Print the list

   Traverse from head, print each node's data followed by ' -> ', end with 'null'.

3. Final Answer:

   Code snippet in Option A correctly inserts and prints the list -> Option A

4. Quick Check:

   New node points to old head, head updated [OK]

Hint: New node next = old head; head = new node; print from head [OK]

Common Mistakes:

• Linking new node incorrectly
• Not updating head to new node
• Appending new node at end instead of beginning