Why Linked List Exists and What Problem It Solves in DSA Java - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a linked list node with a value.

DSA Java

class Node {
    int value;
    Node next;

    Node(int val) {
        this.value = [1];
        this.next = null;
    }
}

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

Aval

Bvalue

Cnext

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using the field name instead of the parameter name.

Assigning null instead of the value.

The constructor assigns the passed value val to the node's value field.

2fill in blank

medium

Complete the code to add a new node at the end of the linked list.

DSA Java

void append(int newValue) {
    Node newNode = new Node(newValue);
    if (head == null) {
        head = newNode;
        return;
    }
    Node current = head;
    while (current.[1] != null) {
        current = current.next;
    }
    current.next = newNode;
}

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

Anext

Bvalue

Chead

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Checking the value instead of the next pointer.

Using an undefined variable like prev.

We traverse the list by moving through the next pointers until we find the last node.

3fill in blank

hard

Fix the error in the code to correctly insert a node at the beginning of the linked list.

DSA Java

void insertAtBeginning(int newValue) {
    Node newNode = new Node(newValue);
    newNode.next = [1];
    head = newNode;
}

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

Anext

BnewNode

Cnull

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Setting newNode.next to null loses the rest of the list.

Assigning newNode.next to itself causes a cycle.

The new node's next should point to the current head before updating head to the new node.

4fill in blank

hard

Fill both blanks to complete the method that counts the number of nodes in the linked list.

DSA Java

int countNodes() {
    int count = 0;
    Node current = head;
    while (current [1] null) {
        count[2];
        current = current.next;
    }
    return count;
}

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

A!=

B==

C++

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Using == null in the loop condition causes no iteration.

Decrementing count instead of incrementing.

The loop runs while current is not null, and count is incremented each time.

5fill in blank

hard

Fill all three blanks to complete the method that searches for a value in the linked list and returns true if found.

DSA Java

boolean contains(int target) {
    Node current = head;
    while (current != null) {
        if (current.[1] == [2]) {
            return [3];
        }
        current = current.next;
    }
    return false;
}

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

Avalue

Btarget

Ctrue

Dfalse

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing the wrong field or variable.

Returning false inside the loop instead of after.

We check if the current node's value equals the target and return true if found.

Practice

1. Why do we use a linked list instead of an array in some cases?

easy

A. Because linked lists allow easy insertion and deletion without shifting elements

B. Because linked lists use less memory than arrays always

C. Because linked lists provide faster access to elements by index

D. Because linked lists store elements in contiguous memory locations

Solution

1. Step 1: Understand array limitations

   Arrays store elements in contiguous memory, so inserting or deleting elements requires shifting others, which is slow.

2. Step 2: Understand linked list advantages

   Linked lists store elements as nodes linked by pointers, so insertion and deletion can be done by changing pointers without shifting.

3. Final Answer:

   Because linked lists allow easy insertion and deletion without shifting elements -> Option A

4. Quick Check:

   Linked list insertion/deletion is easier than arrays [OK]

Hint: Linked lists avoid shifting elements on insert/delete [OK]

Common Mistakes:

• Thinking linked lists always use less memory
• Believing linked lists provide faster index access
• Assuming linked lists store data contiguously

2. Which of the following is the correct way to declare a node class for a singly linked list in Java?

easy

A. class Node { int data; Node next; Node previous; }

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

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

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

Solution

1. Step 1: Identify singly linked list node structure

   A singly linked list node stores data and a reference to the next node only.

2. Step 2: Check options for correct fields

   class Node { int data; Node next; } has int data and Node next, matching singly linked list structure.

3. Final Answer:

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

4. Quick Check:

   Singly linked list node has data and next pointer [OK]

Hint: Singly linked list node has only next pointer [OK]

Common Mistakes:

• Using previous pointer in singly linked list node
• Using int type for next pointer
• Confusing doubly linked list node with singly linked list node

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);
    head.next = new Node(4);

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

medium

A. 1 -> 2 -> 3 -> null

B. 1 -> null

C. 1 -> 4 -> 3 -> null

D. 1 -> 4 -> null

Solution

1. Step 1: Trace node creation and linking

   Initially, head points to node with 1, next points to 2, then 3. Then head.next is reassigned to new node with 4, so node 2 and 3 are disconnected.

2. Step 2: Trace output loop

   Starting at head (1), next is 4, then next is null. So output is 1 -> 4 -> null.

3. Final Answer:

   1 -> 4 -> null -> Option D

4. Quick Check:

   Reassigning head.next breaks old links [OK]

Hint: Reassigning next breaks old chain links [OK]

Common Mistakes:

• Assuming old nodes remain linked after reassignment
• Printing nodes beyond reassigned next
• Confusing node references and values

4. Find the error in the following code snippet that tries to insert a new node at the beginning of a linked list:

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

public class LinkedList {
  Node head;

  void insertAtBeginning(int value) {
    Node newNode = new Node(value);
    newNode.next = head.next;
    head = newNode;
  }
}

medium

A. newNode.next should be assigned head, not head.next

B. head should be assigned newNode.next

C. Node constructor is missing

D. No error, code is correct

Solution

1. Step 1: Understand insertion at beginning logic

   To insert at beginning, newNode.next should point to current head, then head updated to newNode.

2. Step 2: Identify mistake in code

   Code assigns newNode.next = head.next, skipping the current head node, losing the first node.

3. Final Answer:

   newNode.next should be assigned head, not head.next -> Option A

4. Quick Check:

   Insert at beginning links newNode.next to head [OK]

Hint: Link new node to head, not head.next [OK]

Common Mistakes:

• Assigning newNode.next to head.next instead of head
• Not updating head to newNode
• Confusing next pointer assignments

5. You want to store a list of tasks where you frequently add and remove tasks from the middle of the list. Which data structure is best and why?

hard

A. Array, because it allows fast random access

B. Linked list, because it allows efficient insertion and deletion anywhere

C. Stack, because it supports last-in-first-out operations

D. Queue, because it supports first-in-first-out operations

Solution

1. Step 1: Analyze operation requirements

   Adding and removing tasks from the middle requires efficient insertion and deletion at arbitrary positions.

2. Step 2: Compare data structures

   Arrays require shifting elements for middle insert/delete, stacks and queues restrict insertion/removal to ends, linked lists allow efficient insert/delete anywhere by pointer changes.

3. Final Answer:

   Linked list, because it allows efficient insertion and deletion anywhere -> Option B

4. Quick Check:

   Linked list suits frequent middle insert/delete [OK]

Hint: Linked lists excel at middle insert/delete [OK]

Common Mistakes:

• Choosing array for frequent middle insert/delete
• Confusing stack or queue with general list
• Ignoring shifting cost in arrays