Insert at Middle Specific Position in DSA Java - Build from Scratch

Insert at Middle Specific Position

📖 Scenario: Imagine you are managing a queue of customers waiting for service. Sometimes, a new customer arrives and wants to join the queue at a specific position in the middle, not just at the front or the end.

🎯 Goal: You will create a linked list representing the queue and write code to insert a new customer at a specific middle position.

📋 What You'll Learn

Create a linked list with initial customers

Define the position where the new customer will be inserted

Write code to insert the new customer at the given position

Print the linked list after insertion

💡 Why This Matters

🌍 Real World

Queues are common in customer service, task scheduling, and event handling where sometimes new entries need to be inserted in the middle based on priority.

💼 Career

Understanding how to manipulate linked lists and insert elements at specific positions is essential for software developers working with dynamic data structures and real-time systems.

1

Create the initial linked list

Create a linked list called queue and add these exact customers in order: "Alice", "Bob", "Charlie", "Diana".

DSA Java

// Create a LinkedList called queue and add "Alice", "Bob", "Charlie", "Diana"
// Your code here

Hint

Use LinkedList<String> and add() method to add customers.

2

Define the insertion position and new customer

Create an integer variable called position and set it to 2. Create a string variable called newCustomer and set it to "Eve".

DSA Java

import java.util.LinkedList;

public class Main {
    public static void main(String[] args) {
        LinkedList queue = new LinkedList();
        queue.add("Alice");
        queue.add("Bob");
        queue.add("Charlie");
        queue.add("Diana");

        // Create int position = 2 and String newCustomer = "Eve"
        // Your code here
    }
}

Hint

Use int position = 2; and String newCustomer = "Eve";.

3

Insert the new customer at the specific position

Use the add method on queue to insert newCustomer at position.

DSA Java

        queue.add(position, newCustomer);
        // Print the queue

Hint

Use queue.add(position, newCustomer); to insert at the middle.

4

Print the queue after insertion

Use System.out.println(queue); to print the linked list queue after insertion.

DSA Java

Hint

Use System.out.println(queue); to display the list.

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