Node Structure and Pointer Design in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Linked List Node Traversal

What is the printed output of the following Java code that creates and traverses a singly linked list?

DSA Java

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

public class Main {
    public static void main(String[] args) {
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);

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

A 1 -> 2 -> 3 -> null

B 1 -> 3 -> 2 -> null

C 3 -> 2 -> 1 -> null

D null -> 1 -> 2 -> 3

💡 Hint

Think about how nodes are linked and traversed from head to next.

The code creates nodes with data 1, 2, and 3 linked in order. Traversal prints data from head to last node, so output is 1 -> 2 -> 3 -> null.

❓ Predict Output

intermediate

Output of Doubly Linked List Node Insertion

What is the printed output after inserting a new node with value 4 at the end of this doubly linked list?

DSA Java

class Node {
    int data;
    Node prev, next;
    Node(int data) {
        this.data = data;
        this.prev = null;
        this.next = null;
    }
}

public class Main {
    public static void main(String[] args) {
        Node head = new Node(1);
        Node second = new Node(2);
        Node third = new Node(3);

        head.next = second;
        second.prev = head;
        second.next = third;
        third.prev = second;

        Node newNode = new Node(4);
        third.next = newNode;
        newNode.prev = third;

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

A 4 <-> 1 <-> 2 <-> 3 <-> null

B 1 <-> 2 <-> 3 <-> 4 <-> null

C 1 <-> 2 <-> 4 <-> 3 <-> null

D null <-> 1 <-> 2 <-> 3 <-> 4

💡 Hint

New node is added after the last node, so it appears at the end.

The new node with data 4 is linked after node 3. Traversal from head prints nodes in order: 1 <-> 2 <-> 3 <-> 4 <-> null.

🔧 Debug

advanced

Identify the Error in Circular Linked List Node Setup

What error will occur when running this code that tries to create a circular singly linked list?

DSA Java

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

public class Main {
    public static void main(String[] args) {
        Node head = new Node(1);
        Node second = new Node(2);
        Node third = new Node(3);

        head.next = second;
        second.next = third;
        third.next = head;

        Node current = head;
        for (int i = 0; i < 5; i++) {
            System.out.print(current.data + " -> ");
            current = current.next;
        }
        System.out.print("null");
    }
}

A No error; output: 1 -> 2 -> 3 -> 1 -> 2 -> null

B Infinite loop causing program to hang

C NullPointerException at current.next

D ArrayIndexOutOfBoundsException in for loop

💡 Hint

The for loop limits traversal to 5 nodes even though list is circular.

The code creates a circular list and prints 5 nodes. The for loop prevents infinite loop. Output is 1 -> 2 -> 3 -> 1 -> 2 -> null.

❓ Predict Output

advanced

Output of Node Pointer Reassignment in Linked List

What is the output after reassigning pointers in this singly linked list?

DSA Java

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

public class Main {
    public static void main(String[] args) {
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);

        Node temp = head.next; // node with data 2
        head.next = head.next.next; // head.next points to node with data 3
        head.next.next = temp; // node 3 points to node 2
        temp.next = head; // node 2 points back to head

        Node current = head;
        for (int i = 0; i < 4; i++) {
            System.out.print(current.data + " -> ");
            current = current.next;
        }
        System.out.print("null");
    }
}

A 1 -> 3 -> null -> null -> null

B 1 -> 3 -> null

C 1 -> 3 -> 2 -> 1 -> null

D 1 -> 2 -> 3 -> null

💡 Hint

Notice how temp.next points back to head creating a cycle.

After pointer changes, list is: 1 -> 3 -> 2 -> 1 -> ... The for loop prints 4 nodes, so output is 1 -> 3 -> 2 -> 1 -> null.

🧠 Conceptual

expert

Number of Pointers in a Node for Different Data Structures

Which data structure node requires exactly two pointers to other nodes for its basic operation?

A Singly linked list node

B Stack implemented with array

C Binary tree node

D Doubly linked list node

💡 Hint

Think about how many directions each node connects to in the structure.

A doubly linked list node has two pointers: one to the previous node and one to the next node. Singly linked list has one pointer, binary tree nodes have two child pointers but also a parent pointer is optional, stack with array uses no pointers.

Practice

1. What does the next pointer in a singly linked list node represent?

easy

A. It points to the next node in the list

B. It stores the data of the node

C. It points to the previous node in the list

D. It holds the size of the list

Solution

1. Step 1: Understand the role of next pointer

   The next pointer in a singly linked list node is designed to link the current node to the following node in the list.

2. Step 2: Differentiate from other pointers

   It does not store data or point backwards; those roles belong to other parts of the node or list structure.

3. Final Answer:

   It points to the next node in the list -> Option A

4. Quick Check:

   next pointer = next node address [OK]

Hint: Remember: next always moves forward in the list [OK]

Common Mistakes:

• Confusing next with prev pointer
• Thinking next stores data
• Assuming next points to the list head

2. Which of the following is the correct way to declare a node class with a pointer to the next node in Java?

easy

A. class Node { int data; int nextNode; }

B. class Node { int data; Node next; }

C. class Node { int data; int next; }

D. class Node { int data; Node prev; }

Solution

1. Step 1: Check pointer declaration syntax

   In Java, to link nodes, the pointer must be of the same class type, here Node next; is correct.

2. Step 2: Verify naming and types

   class Node { int data; int nextNode; } incorrectly uses int nextNode; which is wrong type. class Node { int data; Node prev; } uses prev which is for doubly linked lists. class Node { int data; int next; } incorrectly uses int next; which is wrong type. class Node { int data; Node next; } uses Node next; which is the standard pointer to the next node.

3. Final Answer:

   class Node { int data; Node next; } -> Option B

4. Quick Check:

   Pointer type = Node, name = next [OK]

Hint: Pointer to next node must be same class type named 'next' [OK]

Common Mistakes:

• Using primitive types instead of Node for pointer
• Confusing next with prev
• Incorrect variable naming causing confusion

3. What will be the output of the following Java code snippet?

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

public class Test {
  public static void main(String[] args) {
    Node head = new Node(1);
    head.next = new Node(2);
    head.next.next = new Node(3);
    System.out.print(head.data + " -> " + head.next.data + " -> " + head.next.next.data + " -> null");
  }
}

medium

A. Compilation error

B. 1 -> 3 -> 2 -> null

C. 3 -> 2 -> 1 -> null

D. 1 -> 2 -> 3 -> null

Solution

1. Step 1: Trace node creation and linking

   Node with data 1 is head, its next points to node with data 2, which points to node with data 3.

2. Step 2: Print statement output

   The print outputs the data in order: 1, then 2, then 3, followed by '-> null'.

3. Final Answer:

   1 -> 2 -> 3 -> null -> Option D

4. Quick Check:

   Linked list order = 1, 2, 3 [OK]

Hint: Follow next pointers step-by-step [OK]

Common Mistakes:

• Mixing node data order
• Assuming reversed list
• Expecting syntax errors where none exist

4. Identify the error in the following node insertion code for a singly linked list:

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

public class Test {
  public static void main(String[] args) {
    Node head = new Node(1);
    Node newNode = new Node(2);
    newNode.next = head;
    head = newNode.next;
    System.out.print(head.data);
  }
}

medium

A. Node constructor is missing

B. newNode.next should be null

C. head is incorrectly assigned, losing newNode reference

D. No error, code runs fine

Solution

1. Step 1: Analyze pointer assignments

   Initially, newNode.next = head; correctly points newNode to old head. But then head = newNode.next; sets head back to old head, losing newNode.

2. Step 2: Identify correct assignment

   To insert newNode at front, head should be assigned to newNode, not newNode.next.

3. Final Answer:

   head is incorrectly assigned, losing newNode reference -> Option C

4. Quick Check:

   head must point to newNode after insertion [OK]

Hint: After insertion, head must point to new node [OK]

Common Mistakes:

• Assigning head to newNode.next instead of newNode
• Confusing pointer directions
• Forgetting to update head after insertion

5. Given a singly linked list node structure with int data and Node next, how would you design a method to insert a new node with value val after a node with a specific value target? Assume the list may not contain target. What is the best approach?

hard

A. Traverse the list to find target, then insert new node after it if found

B. Insert new node at the head regardless of target

C. Replace the data of the node containing target with val

D. Create a new list starting from val and discard old list

Solution

1. Step 1: Understand insertion after a target node

   To insert after a node with value target, we must first find that node by traversing the list.

2. Step 2: Insert new node if target found

   Once found, create new node with val, set its next to target's next, then update target's next to new node.

3. Step 3: Handle case when target not found

   If target is not found, no insertion occurs to keep list intact.

4. Final Answer:

   Traverse the list to find target, then insert new node after it if found -> Option A

5. Quick Check:

   Find target node first, then insert after it [OK]

Hint: Search target node first, then link new node after it [OK]

Common Mistakes:

• Inserting without searching for target
• Replacing data instead of inserting node
• Discarding existing list accidentally