Insert at Middle Specific Position in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does 'Insert at Middle Specific Position' mean in a linked list?

It means adding a new node at a given position inside the list, not at the start or end, but somewhere in the middle.

beginner

What is the first step to insert a node at a specific middle position in a singly linked list?

Traverse the list to reach the node just before the desired position where the new node will be inserted.

intermediate

Why do we need to update the 'next' pointer of the new node and the previous node during insertion?

To link the new node correctly into the list, the previous node's 'next' should point to the new node, and the new node's 'next' should point to the next node in the list.

intermediate

What happens if you try to insert at a position greater than the list length + 1?

The insertion should fail or add the node at the end, depending on implementation, because the position is invalid or beyond the list size.

beginner

In Java, what class structure is commonly used to represent a node in a singly linked list?

A class with two fields: one for the data (like int or String) and one for the reference to the next node (Node next).

What is the correct order of steps to insert a node at position 3 in a singly linked list?

A Create new node, traverse to position 3, adjust pointers

B Adjust pointers, create new node, traverse to position 3

C Traverse to position 2, create new node, adjust pointers

D Traverse to position 3, create new node, adjust pointers

You first reach the node before the insertion point (position 2), then create the new node, and finally adjust pointers to insert.

If you want to insert at the start of a singly linked list, which position do you use?

A Position 1

B Position -1

C Position 2

D Position 0

Positions usually start at 1 for the first node, so position 1 means the start.

What pointer does the new node's 'next' field point to after insertion?

A The previous node

B The node currently at the insertion position

C Null

D The head node

The new node points to the node that was originally at the insertion position.

What happens if you insert a node at position 1 in an empty singly linked list?

A The new node becomes the head

B Insertion fails

C The list remains empty

D The new node points to itself

Inserting at position 1 in an empty list makes the new node the head.

Which of these is NOT needed to insert a node at a middle position?

A Traversing the list

B Creating a new node

C Updating pointers

D Sorting the list

Sorting is not required for insertion at a specific position.

Explain step-by-step how to insert a new node at position 4 in a singly linked list.

Think about how pointers change to keep the list connected.

What are the edge cases to consider when inserting a node at a specific position in a linked list?

Consider list size and valid positions.

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