Delete from End of Doubly Linked List in DSA Go - 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 Go

package main

import "fmt"

type Node struct {
    data int
    prev *Node
    next *Node
}

type DoublyLinkedList struct {
    head *Node
}

func (dll *DoublyLinkedList) append(data int) {
    newNode := &Node{data: data}
    if dll.head == nil {
        dll.head = newNode
        return
    }
    temp := dll.head
    for temp.next != nil {
        temp = temp.next
    }
    temp.next = newNode
    newNode.prev = temp
}

func (dll *DoublyLinkedList) deleteFromEnd() {
    if dll.head == nil {
        return
    }
    if dll.head.next == nil {
        dll.head = nil
        return
    }
    temp := dll.head
    for temp.next != nil {
        temp = temp.next
    }
    temp.prev.next = nil
}

func (dll *DoublyLinkedList) printList() {
    if dll.head == nil {
        fmt.Println("null")
        return
    }
    temp := dll.head
    for temp != nil {
        fmt.Printf("%d", temp.data)
        if temp.next != nil {
            fmt.Printf(" -> ")
        }
        temp = temp.next
    }
    fmt.Println(" -> null")
}

func main() {
    dll := DoublyLinkedList{}
    dll.append(1)
    dll.append(2)
    dll.append(3)
    dll.deleteFromEnd()
    dll.printList()
}

A 1 -> 2 -> null

B 1 -> 2 -> 3 -> null

C 2 -> 3 -> null

D null

💡 Hint

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

The list initially is 1 -> 2 -> 3 -> null. After deleting the last node (3), the list becomes 1 -> 2 -> 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 Go

package main

import "fmt"

type Node struct {
    data int
    prev *Node
    next *Node
}

type DoublyLinkedList struct {
    head *Node
}

func (dll *DoublyLinkedList) append(data int) {
    newNode := &Node{data: data}
    if dll.head == nil {
        dll.head = newNode
        return
    }
    temp := dll.head
    for temp.next != nil {
        temp = temp.next
    }
    temp.next = newNode
    newNode.prev = temp
}

func (dll *DoublyLinkedList) deleteFromEnd() {
    if dll.head == nil {
        return
    }
    if dll.head.next == nil {
        dll.head = nil
        return
    }
    temp := dll.head
    for temp.next != nil {
        temp = temp.next
    }
    temp.prev.next = nil
}

func (dll *DoublyLinkedList) printList() {
    if dll.head == nil {
        fmt.Println("null")
        return
    }
    temp := dll.head
    for temp != nil {
        fmt.Printf("%d", temp.data)
        if temp.next != nil {
            fmt.Printf(" -> ")
        }
        temp = temp.next
    }
    fmt.Println(" -> null")
}

func main() {
    dll := DoublyLinkedList{}
    dll.append(10)
    dll.deleteFromEnd()
    dll.printList()
}

A 10

B 10 -> null

C Error: nil pointer dereference

D null

💡 Hint

Deleting the only node should leave the list empty.

When the list has only one node, deleting from end sets the head to nil, resulting in an empty list printed as null.

🔧 Debug

advanced

Identify the error in deleteFromEnd method

What error will this code produce when deleting from an empty doubly linked list?

DSA Go

func (dll *DoublyLinkedList) deleteFromEnd() {
    temp := dll.head
    for temp.next != nil {
        temp = temp.next
    }
    temp.prev.next = nil
}

A compilation error: temp not declared

B no error, deletes last node correctly

C runtime error: invalid memory address or nil pointer dereference

D syntax error: missing return statement

💡 Hint

Check what happens if dll.head is nil before accessing temp.next.

If dll.head is nil, temp is nil. Accessing temp.next causes a nil pointer dereference runtime error.

❓ Predict Output

advanced

Output after multiple deletions from end

What is the printed state of the doubly linked list after deleting from end twice?

DSA Go

package main

import "fmt"

type Node struct {
    data int
    prev *Node
    next *Node
}

type DoublyLinkedList struct {
    head *Node
}

func (dll *DoublyLinkedList) append(data int) {
    newNode := &Node{data: data}
    if dll.head == nil {
        dll.head = newNode
        return
    }
    temp := dll.head
    for temp.next != nil {
        temp = temp.next
    }
    temp.next = newNode
    newNode.prev = temp
}

func (dll *DoublyLinkedList) deleteFromEnd() {
    if dll.head == nil {
        return
    }
    if dll.head.next == nil {
        dll.head = nil
        return
    }
    temp := dll.head
    for temp.next != nil {
        temp = temp.next
    }
    temp.prev.next = nil
}

func (dll *DoublyLinkedList) printList() {
    if dll.head == nil {
        fmt.Println("null")
        return
    }
    temp := dll.head
    for temp != nil {
        fmt.Printf("%d", temp.data)
        if temp.next != nil {
            fmt.Printf(" -> ")
        }
        temp = temp.next
    }
    fmt.Println(" -> null")
}

func main() {
    dll := DoublyLinkedList{}
    dll.append(5)
    dll.append(10)
    dll.append(15)
    dll.deleteFromEnd()
    dll.deleteFromEnd()
    dll.printList()
}

A 5 -> null

B 5 -> 10 -> null

C 10 -> 15 -> null

D null

💡 Hint

Each deleteFromEnd removes the last node, so two deletions remove 15 and then 10.

Starting list: 5 -> 10 -> 15 -> null. After first deletion: 5 -> 10 -> null. After second deletion: 5 -> null.

🧠 Conceptual

expert

Why is prev pointer important in deleteFromEnd for doubly linked list?

Which statement best explains why the prev pointer is used when deleting the last node in a doubly linked list?

A To find the head of the list quickly without traversal

B To update the next pointer of the previous node to nil, effectively removing the last node

C To reverse the list after deletion automatically

D To avoid using the next pointer which is unreliable

💡 Hint

Think about how to remove the last node without losing the rest of the list.

The prev pointer allows access to the node before the last node so its next pointer can be set to nil, removing the last node cleanly.

Practice

1. What happens to the tail pointer when you delete the last node from a doubly linked list?

easy

A. The tail pointer becomes null immediately.

B. The tail pointer moves to the next node.

C. The tail pointer does not change.

D. The tail pointer moves to the previous node.

Solution

1. Step 1: Understand the tail pointer role

   The tail pointer points to the last node in the doubly linked list.

2. Step 2: Effect of deleting the last node

   When the last node is deleted, the tail pointer must move to the previous node to remain valid.

3. Final Answer:

   The tail pointer moves to the previous node. -> Option D

4. Quick Check:

   Tail moves backward after deletion [OK]

Hint: Tail always points to last node; update to previous on delete [OK]

Common Mistakes:

• Not updating tail pointer after deletion
• Setting tail to next node which is null
• Leaving tail pointer unchanged

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

easy

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

B. dll.tail = dll.tail.prev; dll.tail.next = nil

C. dll.head = dll.head.next; dll.head.prev = nil

D. dll.head = dll.tail; dll.tail = nil

Solution

1. Step 1: Identify correct pointer updates for tail deletion

   Deleting last node means tail moves to previous node and new tail's next is nil.

2. Step 2: Check code correctness

   dll.tail = dll.tail.prev; dll.tail.next = nil sets tail to previous node and sets tail.next to nil, which is correct.

3. Final Answer:

   dll.tail = dll.tail.prev; dll.tail.next = nil -> Option B

4. Quick Check:

   Tail moves back, next cleared [OK]

Hint: Tail moves to prev; set tail.next to nil [OK]

Common Mistakes:

• Moving tail forward instead of backward
• Not setting tail.next to nil
• Changing head pointer instead of tail

3. What is the output of the following Go code that deletes the last node from a doubly linked list?

package main
import "fmt"

type Node struct {
    data int
    prev *Node
    next *Node
;
}

type DoublyLinkedList struct {
    head *Node
    tail *Node
;
}

func (dll *DoublyLinkedList) DeleteFromEnd() {
    if dll.tail == nil {
        return
    }
    if dll.tail.prev == nil {
        dll.head = nil
        dll.tail = nil
        return
    }
    dll.tail = dll.tail.prev
    dll.tail.next = nil
}

func (dll *DoublyLinkedList) Print() {
    current := dll.head
    for current != nil {
        fmt.Print(current.data, " -> ")
        current = current.next
    }
    fmt.Println("null")
}

func main() {
    n1 := &Node{data: 1}
    n2 := &Node{data: 2}
    n3 := &Node{data: 3}
    n1.next = n2
    n2.prev = n1
    n2.next = n3
    n3.prev = n2
    dll := DoublyLinkedList{head: n1, tail: n3}
    dll.DeleteFromEnd()
    dll.Print()
}

medium

A. 1 -> 2 -> null

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

C. 2 -> 3 -> null

D. null

Solution

1. Step 1: Initial list and deletion

   List is 1 -> 2 -> 3 -> null. DeleteFromEnd removes last node (3).

2. Step 2: After deletion

   Tail moves to node with data 2, tail.next set to nil. List becomes 1 -> 2 -> null.

3. Final Answer:

   1 -> 2 -> null -> Option A

4. Quick Check:

   Last node removed, tail updated [OK]

Hint: Deleting end removes last node, tail moves back [OK]

Common Mistakes:

• Not updating tail.next to nil causing dangling pointer
• Printing old list without deletion
• Confusing head and tail pointers

4. The following Go code attempts to delete the last node from a doubly linked list but causes a runtime panic. What is the error?

func (dll *DoublyLinkedList) DeleteFromEnd() {
    if dll.tail == nil {
        return
    }
    dll.tail = dll.tail.prev
    dll.tail.next = nil
}

medium

A. It should delete from the head, not the tail.

B. It incorrectly sets tail.next to nil before moving tail.

C. It does not handle the case when the list has only one node.

D. It should set head to nil instead of tail.

Solution

1. Step 1: Analyze edge case for single node list

   If list has one node, tail.prev is nil, so after tail = tail.prev, tail is nil.

2. Step 2: Cause of panic

   Next line tries to access tail.next but tail is nil, causing runtime panic.

3. Final Answer:

   It does not handle the case when the list has only one node. -> Option C

4. Quick Check:

   Missing single-node check causes nil pointer panic [OK]

Hint: Check for single-node list before updating tail [OK]

Common Mistakes:

• Not checking if tail.prev is nil before access
• Assuming list always has multiple nodes
• Ignoring head pointer update when list becomes empty

5. You have a doubly linked list with nodes: 10 <-> 20 <-> 30 <-> 40. After deleting the last node twice using the correct delete from end method, what will be the printed list?

hard

A. 10 -> 20 -> null

B. 10 -> 20 -> 30 -> null

C. 20 -> 30 -> null

D. 10 -> 30 -> null

Solution

1. Step 1: Initial list and first deletion

   List is 10 <-> 20 <-> 30 <-> 40. First deletion removes 40, list becomes 10 <-> 20 <-> 30.

2. Step 2: Second deletion

   Second deletion removes 30, list becomes 10 <-> 20.

3. Step 3: Print list

   Printing from head: 10 -> 20 -> null.

4. Final Answer:

   10 -> 20 -> null -> Option A

5. Quick Check:

   Two deletions remove last two nodes [OK]

Hint: Each delete removes last node, tail moves back [OK]

Common Mistakes:

• Deleting only once instead of twice
• Confusing node values after deletion
• Not updating tail pointer correctly