Delete from End of Doubly Linked List in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output after deleting last node from doubly linked list

What is the printed state of the doubly linked list after deleting the last node?

DSA Java

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

class DoublyLinkedList {
    Node head;

    void append(int data) {
        Node newNode = new Node(data);
        if (head == null) {
            head = newNode;
            return;
        }
        Node last = head;
        while (last.next != null) last = last.next;
        last.next = newNode;
        newNode.prev = last;
    }

    void deleteFromEnd() {
        if (head == null) return;
        if (head.next == null) {
            head = null;
            return;
        }
        Node last = head;
        while (last.next != null) last = last.next;
        last.prev.next = null;
    }

    void printList() {
        Node temp = head;
        while (temp != null) {
            System.out.print(temp.data + " -> ");
            temp = temp.next;
        }
        System.out.print("null\n");
    }

    public static void main(String[] args) {
        DoublyLinkedList dll = new DoublyLinkedList();
        dll.append(10);
        dll.append(20);
        dll.append(30);
        dll.deleteFromEnd();
        dll.printList();
    }
}

A 10 -> 20 -> null

B null

C 20 -> 30 -> null

D 10 -> 20 -> 30 -> null

💡 Hint

Deleting from end removes the last node and updates the previous node's next pointer to null.

The list initially is 10 -> 20 -> 30 -> null. After deleting the last node (30), the list becomes 10 -> 20 -> null.

❓ Predict Output

intermediate

Output after deleting last node from single-node doubly linked list

What is the printed state of the doubly linked list after deleting the last node when the list has only one node?

DSA Java

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

class DoublyLinkedList {
    Node head;

    void append(int data) {
        Node newNode = new Node(data);
        if (head == null) {
            head = newNode;
            return;
        }
        Node last = head;
        while (last.next != null) last = last.next;
        last.next = newNode;
        newNode.prev = last;
    }

    void deleteFromEnd() {
        if (head == null) return;
        if (head.next == null) {
            head = null;
            return;
        }
        Node last = head;
        while (last.next != null) last = last.next;
        last.prev.next = null;
    }

    void printList() {
        Node temp = head;
        while (temp != null) {
            System.out.print(temp.data + " -> ");
            temp = temp.next;
        }
        System.out.print("null\n");
    }

    public static void main(String[] args) {
        DoublyLinkedList dll = new DoublyLinkedList();
        dll.append(100);
        dll.deleteFromEnd();
        dll.printList();
    }
}

A null

B 100 -> null

C Error

D 100

💡 Hint

Deleting the only node should result in an empty list.

Since the list has only one node, deleting from end sets head to null, resulting in an empty list.

🔧 Debug

advanced

Identify the error in deleteFromEnd method

What error will occur when running this deleteFromEnd method on a doubly linked list with multiple nodes?

DSA Java

void deleteFromEnd() {
    if (head == null) return;
    Node last = head;
    while (last.next != null) last = last.next;
    last.next = null;
}

A No error, works correctly

B The last node is not deleted

C NullPointerException

D Infinite loop

💡 Hint

Consider what happens to the previous node's next pointer.

The method tries to set last.next = null, but last.next is already null. It does not update the previous node's next pointer, so the list still points to the last node. Accessing last.prev.next causes NullPointerException if last.prev is null or inconsistent state.

🧠 Conceptual

advanced

Why update previous node's next pointer when deleting last node?

In a doubly linked list, why must we update the previous node's next pointer to null when deleting the last node?

A To update the previous node's prev pointer

B To make the list circular

C To remove the reference to the deleted node and avoid memory leaks

D To update the head pointer to the last node

💡 Hint

Think about what happens if the previous node still points to the deleted node.

If the previous node's next pointer is not set to null, it still points to the deleted node, causing dangling references and potential memory leaks.

❓ Predict Output

expert

Output after multiple deletions from end of doubly linked list

What is the printed state of the doubly linked list after deleting the last node three times in a row?

DSA Java

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

class DoublyLinkedList {
    Node head;

    void append(int data) {
        Node newNode = new Node(data);
        if (head == null) {
            head = newNode;
            return;
        }
        Node last = head;
        while (last.next != null) last = last.next;
        last.next = newNode;
        newNode.prev = last;
    }

    void deleteFromEnd() {
        if (head == null) return;
        if (head.next == null) {
            head = null;
            return;
        }
        Node last = head;
        while (last.next != null) last = last.next;
        last.prev.next = null;
    }

    void printList() {
        Node temp = head;
        while (temp != null) {
            System.out.print(temp.data + " -> ");
            temp = temp.next;
        }
        System.out.print("null\n");
    }

    public static void main(String[] args) {
        DoublyLinkedList dll = new DoublyLinkedList();
        dll.append(1);
        dll.append(2);
        dll.append(3);
        dll.append(4);
        dll.append(5);
        dll.deleteFromEnd();
        dll.deleteFromEnd();
        dll.deleteFromEnd();
        dll.printList();
    }
}

A 1 -> 2 -> 3 -> null

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

C null

D 1 -> 2 -> null

💡 Hint

Each deletion removes the last node, so after three deletions from a list of 5 nodes, 2 nodes remain.

Starting with 1 -> 2 -> 3 -> 4 -> 5 -> null, deleting last node thrice removes 5, 4, and 3, leaving 1 -> 2 -> null.

Practice

1. What happens when you delete the last node from a doubly linked list that has multiple nodes?

easy

A. The second last node becomes the new tail and its next pointer is set to null.

B. The head pointer is set to null.

C. The list becomes circular.

D. The first node is deleted instead.

Solution

1. Step 1: Identify the last node and its previous node

   In a doubly linked list, the last node's previous node is the second last node.

2. Step 2: Update the second last node's next pointer

   Set the second last node's next pointer to null, making it the new tail.

3. Final Answer:

   The second last node becomes the new tail and its next pointer is set to null. -> Option A

4. Quick Check:

   Deleting last node updates tail's next to null [OK]

Hint: Tail's previous node next pointer becomes null after deletion [OK]

Common Mistakes:

• Setting head to null instead of updating tail
• Not updating the previous node's next pointer
• Deleting the first node by mistake

2. Which of the following Java code snippets correctly deletes the last node from a doubly linked list named dll?

easy

A. dll.tail.next = dll.tail.prev; dll.tail = null;

B. dll.head = dll.head.next; dll.head.prev = null;

C. dll.tail = dll.tail.prev; dll.tail.next = null;

D. dll.head.prev = dll.tail; dll.tail = dll.tail.next;

Solution

1. Step 1: Move tail pointer to previous node

   Set dll.tail to dll.tail.prev to remove last node reference.

2. Step 2: Set new tail's next pointer to null

   Update dll.tail.next to null to mark end of list.

3. Final Answer:

   dll.tail = dll.tail.prev; dll.tail.next = null; -> Option C

4. Quick Check:

   Correct tail update and next pointer null [OK]

Hint: Tail moves back, next pointer set to null [OK]

Common Mistakes:

• Updating head instead of tail
• Setting tail to null without updating pointers
• Confusing next and prev pointers

3. Consider the following Java code for deleting the last node from a doubly linked list:

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

class DoublyLinkedList {
  Node head, tail;

  void deleteFromEnd() {
    if (tail == null) return;
    if (head == tail) {
      head = tail = null;
      return;
    }
    tail = tail.prev;
    tail.next = null;
  }
}

DoublyLinkedList dll = new DoublyLinkedList();
dll.head = new Node(1);
Node second = new Node(2);
dll.head.next = second;
second.prev = dll.head;
dll.tail = second;
dll.deleteFromEnd();
System.out.println(dll.head.data + " -> " + (dll.head.next == null ? "null" : dll.head.next.data));

What is the output?

medium

A. 1 -> 2

B. null -> null

C. 2 -> null

D. 1 -> null

Solution

1. Step 1: Setup list with two nodes 1 and 2

   Head points to node 1, tail points to node 2, linked properly.

2. Step 2: Delete last node (node 2)

   Tail moves to node 1, tail.next set to null, removing node 2.

3. Final Answer:

   1 -> null -> Option D

4. Quick Check:

   After deletion, only node 1 remains with next null [OK]

Hint: After deleting tail, tail.prev.next is null [OK]

Common Mistakes:

• Printing old tail node data
• Not updating tail.next to null
• Confusing head and tail pointers

4. The following code attempts to delete the last node from a doubly linked list but causes a runtime error:

void deleteFromEnd() {
  if (tail == null) return;
  tail = tail.prev;
  tail.next = null;
}

What is the error and how to fix it?

medium

A. Compilation error due to missing semicolon; fix by adding semicolon.

B. NullPointerException when list has one node; fix by checking if tail.prev is null.

C. Logical error deleting head instead of tail; fix by updating head pointer.

D. No error; code works fine.

Solution

1. Step 1: Identify cause of NullPointerException

   If list has one node, tail.prev is null, so tail becomes null, then tail.next causes error.

2. Step 2: Fix by checking single node case

   Add condition: if head == tail, set head and tail to null before updating pointers.

3. Final Answer:

   NullPointerException when list has one node; fix by checking if tail.prev is null. -> Option B

4. Quick Check:

   Check single node case before pointer update [OK]

Hint: Check if tail.prev is null before accessing tail.next [OK]

Common Mistakes:

• Not handling single node list
• Assuming tail.prev always exists
• Ignoring null pointer exceptions

5. You have a doubly linked list with nodes: 10 <-> 20 <-> 30 <-> 40. After deleting the last node twice using the correct deleteFromEnd method, what is the final list?

hard

A. 10 <-> 20

B. 10 <-> 20 <-> 30

C. 20 <-> 30

D. 30 <-> 40

Solution

1. Step 1: Delete last node once

   Deleting last node 40 removes it; list becomes 10 <-> 20 <-> 30.

2. Step 2: Delete last node second time

   Deleting last node 30 removes it; list becomes 10 <-> 20.

3. Final Answer:

   10 <-> 20 -> Option A

4. Quick Check:

   Two deletions remove last two nodes [OK]

Hint: Each deleteFromEnd removes tail node [OK]

Common Mistakes:

• Deleting from head instead of tail
• Not updating tail pointer after deletion
• Confusing node values after deletion