Insert at Beginning Head Insert in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node and insert it at the beginning of the linked list.

DSA Java

Node newNode = new Node(data);
newNode.next = [1];
head = newNode;

Drag options to blanks, or click blank then click option'

Adata

BnewNode

Cnull

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Setting newNode.next to null loses the rest of the list.

Assigning newNode.next to newNode creates a cycle.

The new node's next should point to the current head before updating head to newNode.

2fill in blank

medium

Complete the method to insert a new node at the beginning of the linked list.

DSA Java

public void insertAtBeginning(int data) {
    Node newNode = new Node(data);
    newNode.next = [1];
    head = newNode;
}

Drag options to blanks, or click blank then click option'

Ahead

BnewNode

Cnull

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Setting newNode.next to null loses the rest of the list.

Assigning newNode.next to newNode creates a cycle.

The new node's next should point to the current head before updating head to newNode.

3fill in blank

hard

Fix the error in the insertAtBeginning method to correctly insert a node at the head.

DSA Java

public void insertAtBeginning(int data) {
    Node newNode = new Node(data);
    newNode.next = [1];
    head = newNode;
}

Drag options to blanks, or click blank then click option'

AnewNode

Bhead

Cdata

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Setting newNode.next to newNode causes a cycle.

Setting newNode.next to null loses the rest of the list.

The new node's next must point to the current head to keep the list intact.

4fill in blank

hard

Fill both blanks to create a method that inserts a new node at the beginning and updates the head.

DSA Java

public void insertAtBeginning(int data) {
    Node newNode = new Node([1]);
    newNode.next = [2];
    head = newNode;
}

Drag options to blanks, or click blank then click option'

Adata

Bhead

CnewNode

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using newNode instead of data when creating the node.

Setting newNode.next to null loses the rest of the list.

The new node is created with data, and its next points to the current head before updating head.

5fill in blank

hard

Fill all three blanks to complete the insertAtBeginning method that creates a new node, links it, and updates the head.

DSA Java

public void insertAtBeginning(int [1]) {
    Node newNode = new Node([2]);
    newNode.next = [3];
    head = newNode;
}

Drag options to blanks, or click blank then click option'

Adata

Chead

DnewNode

Attempts:

3 left

💡 Hint

Common Mistakes

Using newNode as method parameter name.

Setting newNode.next to newNode causes a cycle.

The method parameter is data, used to create the new node, whose next points to head before updating head.

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