Insert at Middle Specific Position in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a new node at the given position in the linked list.

DSA Java

Node newNode = new Node(data);
if (position == 0) {
    newNode.next = head;
    head = [1];
}

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

AnewNode

Bhead

Cdata

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning head to newNode.next instead of newNode to head

Forgetting to update head after insertion

To insert at the start, the new node becomes the head.

2fill in blank

medium

Complete the loop to reach the node just before the insertion position.

DSA Java

Node current = head;
for (int i = 0; i < [1] - 1; i++) {
    current = current.next;
}

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

Aposition

Bdata

Chead

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Looping up to position instead of position - 1

Using size or data instead of position

We loop until the node before the desired position to insert.

3fill in blank

hard

Fix the error in linking the new node to the next node.

DSA Java

newNode.next = [1].next;
[2].next = newNode;

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Acurrent

Bhead

CnewNode

Dposition

Attempts:

3 left

💡 Hint

Common Mistakes

Using newNode.next = head.next instead of current.next

Assigning newNode.next = newNode.next causing self-loop

The new node's next should point to current's next, then current's next points to new node.

4fill in blank

hard

Fill both blanks to check if position is valid and handle empty list.

DSA Java

if (position < 0 || position > [1]) {
    throw new IndexOutOfBoundsException();
}
if (head == [2]) {
    head = new Node(data);
    return;
}

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

Asize

Bnull

Chead

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Checking position against head instead of size

Comparing head to data instead of null

Position must be within 0 and size. If list is empty (head is null), create new head.

5fill in blank

hard

Fill all three blanks to complete the insertAtPosition method.

DSA Java

public void insertAtPosition(int data, int position) {
    if (position < 0 || position > [1]) {
        throw new IndexOutOfBoundsException();
    }
    if (position == 0) {
        Node newNode = new Node(data);
        newNode.next = [2];
        head = newNode;
        return;
    }
    Node current = head;
    for (int i = 0; i < position - 1; i++) {
        current = current.next;
    }
    Node newNode = new Node(data);
    newNode.next = current.[3];
    current.next = newNode;
}

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

Asize

Bhead

Cnext

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Not updating head when inserting at position 0

Incorrectly setting newNode.next or current.next

Check position against size, insert at head if position 0, else traverse and insert by adjusting next pointers.

Practice

1. What does inserting a node at a specific middle position in a linked list mean?

easy

A. Adding a new node exactly at the given position counting from the head

B. Adding a new node only at the start of the list

C. Adding a new node only at the end of the list

D. Replacing the node at the given position with a new node

Solution

1. Step 1: Understand position-based insertion

   Inserting at a specific position means placing the new node exactly at that index, shifting others if needed.

2. Step 2: Differentiate from other insertions

   Start or end insertions are special cases, but middle insertion targets a specific index inside the list.

3. Final Answer:

   Adding a new node exactly at the given position counting from the head -> Option A

4. Quick Check:

   Insert at position = Insert node at that index [OK]

Hint: Middle insert means place node exactly at given index [OK]

Common Mistakes:

• Confusing insertion with replacement
• Thinking insertion only happens at start or end
• Ignoring zero-based or one-based indexing

2. Which of the following Java code snippets correctly inserts a new node with value 10 at position 3 in a singly linked list?

easy

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

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

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

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

Solution

1. Step 1: Identify position 3 in zero-based indexing

   Position 3 means after two nodes: head -> next -> next, so newNode.next should point to head.next.next.

2. Step 2: Link newNode correctly

   Set newNode.next to head.next.next, then update head.next.next to newNode to insert at position 3.

3. Final Answer:

   Node newNode = new Node(10); newNode.next = head.next.next; head.next.next = newNode; -> Option C

4. Quick Check:

   Insert at position 3 = link newNode after second node [OK]

Hint: Link new node's next to current node at position, then update previous node [OK]

Common Mistakes:

• Incorrectly setting newNode.next to wrong node
• Not updating previous node's next pointer
• Confusing position indexing

3. What will be the linked list after inserting a node with value 25 at position 2 in the list: 10 -> 20 -> 30 -> 40 -> null?

Node head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);

// Insert 25 at position 2
Node newNode = new Node(25);
Node temp = head;
for (int i = 1; i < 2; i++) {
    temp = temp.next;
}
newNode.next = temp.next;
temp.next = newNode;

medium

A. 25 -> 10 -> 20 -> 30 -> 40 -> null

B. 10 -> 20 -> 30 -> 25 -> 40 -> null

C. 10 -> 20 -> 25 -> 30 -> 40 -> null

D. 10 -> 25 -> 20 -> 30 -> 40 -> null

Solution

1. Step 1: Traverse to node at position 1

   The loop runs once (i=1), so temp points to node with value 20.

2. Step 2: Insert newNode after temp

   newNode.next points to temp.next (node 30), then temp.next points to newNode, inserting 25 after 20.

3. Final Answer:

   10 -> 20 -> 30 -> 25 -> 40 -> null -> Option B

4. Quick Check:

   Insert at position 2 places 25 after 20 [OK]

Hint: Loop to node before position, link new node next, update previous next [OK]

Common Mistakes:

• Looping one extra time causing wrong insertion
• Not updating newNode.next correctly
• Confusing position indexing

4. Identify the error in the following code that tries to insert a node with value 15 at position 1 in a linked list:

Node newNode = new Node(15);
Node temp = head;
for (int i = 0; i < 1; i++) {
    temp = temp.next;
}
newNode.next = temp.next;
temp.next = newNode;

medium

A. Loop moves temp too far, skipping the head node

B. newNode.next is assigned incorrectly

C. temp.next is not updated after insertion

D. No error, code works fine

Solution

1. Step 1: Analyze loop movement

   The loop runs once, moving temp from head to head.next, so temp points to node at position 1, not before it.

2. Step 2: Understand insertion point

   To insert at position 1, temp should point to node before position 1 (head), but here it points to position 1 node, causing incorrect insertion.

3. Final Answer:

   Loop moves temp too far, skipping the head node -> Option A

4. Quick Check:

   Loop should stop before position to insert [OK]

Hint: Loop to node before position, not at position [OK]

Common Mistakes:

• Looping one time too many
• Confusing position indexing
• Not handling insertion at position 0 separately

5. Given a singly linked list with nodes: 5 -> 10 -> 15 -> 20 -> null, write the output after inserting a node with value 12 at position 3 using the following code:

Node newNode = new Node(12);
Node temp = head;
int position = 3;
for (int i = 1; i < position - 1; i++) {
    temp = temp.next;
}
newNode.next = temp.next;
temp.next = newNode;

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

hard

A. 5 -> 10 -> 15 -> 12 -> 20 -> null

B. 12 -> 5 -> 10 -> 15 -> 20 -> null

C. 5 -> 12 -> 10 -> 15 -> 20 -> null

D. 5 -> 10 -> 12 -> 15 -> 20 -> null

Solution

1. Step 1: Loop to node before position 3

   The loop runs with i=1 only, so temp moves from head (5) to next (10), which is node before position 3.

2. Step 2: Insert newNode after temp

   newNode.next points to temp.next (15), then temp.next points to newNode, inserting 12 between 10 and 15.

3. Final Answer:

   5 -> 10 -> 12 -> 15 -> 20 -> null -> Option D

4. Quick Check:

   Insert at position 3 places 12 after 10 [OK]

Hint: Loop to node before position, link new node, update next pointers [OK]

Common Mistakes:

• Looping too many times causing wrong insertion
• Not updating newNode.next correctly
• Confusing position indexing