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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to move to the last node in the doubly linked list.

DSA Go

current := head
for current.[1] != nil {
    current = current.next
}

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

Atail

Bnext

Chead

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' causes backward traversal.

Using 'head' or 'tail' as a pointer is incorrect here.

To reach the last node, we follow the 'next' pointers until nil.

2fill in blank

medium

Complete the code to update the previous node's next pointer when deleting the last node.

DSA Go

if current.prev != nil {
    current.prev.[1] = nil
}

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

Anext

Bprev

Chead

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Setting prev pointer instead of next pointer.

Setting head or tail incorrectly.

We set the previous node's 'next' pointer to nil to remove the last node.

3fill in blank

hard

Fix the error in updating the head pointer when the list has only one node.

DSA Go

if current.prev == nil {
    [1] = nil
}

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

Acurrent

Bprev

Ctail

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Setting current or tail to nil instead of head.

Not updating head at all.

If the only node is deleted, head should be set to nil to indicate an empty list.

4fill in blank

hard

Fill both blanks to correctly delete the last node and update pointers.

DSA Go

if current.prev != nil {
    current.prev.[1] = nil
} else {
    [2] = nil
}

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

Anext

Bprev

Chead

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing next and prev pointers.

Setting tail instead of head when list is empty.

Set previous node's next to nil and if no previous node, set head to nil.

5fill in blank

hard

Fill all three blanks to complete the function that deletes the last node from a doubly linked list.

DSA Go

func deleteEnd(head *Node) *Node {
    if head == nil {
        return nil
    }
    current := head
    for current.[1] != nil {
        current = current.next
    }
    if current.prev != nil {
        current.prev.[2] = nil
    } else {
        [3] = nil
    }
    return head
}

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

Aprev

Bnext

Chead

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev instead of next for traversal.

Setting wrong pointers when deleting the last node.

Traverse using next, set prev node's next to nil, and if no prev node, set head to nil.

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