Insert at Middle Specific Position in DSA Java - Execution Trace

Concept Flow - Insert at Middle Specific Position

Start with head node

↓

Check if position is 1

|Yes↓

Insert at head

No↓

Traverse list to position-1

↓

Create new node

↓

Link new node's next to current node's next

↓

Link current node's next to new node

↓

Insertion complete

↓

End

Start from the head, traverse to the node before the desired position, insert the new node by adjusting links, then finish.

Execution Sample

DSA Java

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

// Insert 99 at position 3 in list 10->20->30->null

Inserts a new node with value 99 at position 3 in the linked list.

Execution Table

Step	Current Node Data	Position Counter	Action	List State
1	10	1	Start at head	10 -> 20 -> 30 -> null
2	10	1	Position != 1, move to next	10 -> 20 -> 30 -> null
3	20	2	Position counter == 2 (pos-1), stop traversal	10 -> 20 -> 30 -> null
4	20	2	Reached position-1 (2), prepare to insert	10 -> 20 -> 30 -> null
5	20	2	Create new node with data 99	10 -> 20 -> 30 -> null
6	20	2	New node next points to current node's next (30)	10 -> 20 -> 30 -> null
7	20	2	20's next points to new node	10 -> 20 -> 99 -> 30 -> null
8	99	Inserted	Insertion complete	10 -> 20 -> 99 -> 30 -> null

💡 Insertion done at position 3, list updated successfully

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 7	Final
current	10	10	20	20	20
positionCounter	1	1	2	2	2
newNode	null	null	null	99	99
list	10->20->30->null	10->20->30->null	10->20->30->null	10->20->99->30->null	10->20->99->30->null

Key Moments - 2 Insights

Why do we stop traversal at position-1 and not at the exact position?

Because to insert a new node at position 'pos', we need to link it after the node at position 'pos-1'. This is shown in execution_table row 4 where we stop at position 2 to insert at position 3.

What happens if position is 1?

If position is 1, we insert the new node at the head. This is checked at the start (concept_flow) and would skip traversal, directly linking new node before current head.

Visual Quiz - 3 Questions

Look at the execution_table, what is the list state after step 7?

A 10 -> 20 -> 30 -> null

B 10 -> 99 -> 20 -> 30 -> null

C 10 -> 20 -> 99 -> 30 -> null

D 99 -> 10 -> 20 -> 30 -> null

💡 Hint

Check the 'List State' column in row 7 of execution_table

At which step is the new node created?

A Step 4

B Step 5

C Step 6

D Step 7

💡 Hint

Look for the action mentioning 'Create new node' in execution_table

If we wanted to insert at position 1, which step would be skipped?

A Step 2 and 3 (traversal steps)

B Step 5 (node creation)

C Step 7 (linking new node)

D Step 8 (completion)

💡 Hint

Refer to concept_flow where position 1 inserts at head without traversal

Concept Snapshot

Insert at Middle Specific Position in Linked List:
- Traverse to node before desired position (pos-1)
- Create new node
- Link new node's next to current node's next
- Link current node's next to new node
- If position is 1, insert at head
- Adjust pointers carefully to avoid losing nodes

Full Transcript

To insert a node at a specific middle position in a linked list, start at the head and check if the position is 1. If yes, insert at the head. Otherwise, traverse the list until you reach the node just before the desired position (position-1). Create a new node with the given data. Set the new node's next pointer to the next node of the current node. Then update the current node's next pointer to point to the new node. This inserts the new node at the correct position without losing any nodes. The execution table shows each step, including traversal, node creation, and pointer updates. The variable tracker shows how the current node, position counter, new node, and list state change during execution.

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