Why Delete from End of Doubly Linked List in DSA Go?

The Big Idea

What if you could remove the last item instantly without checking the whole list?

The Scenario

Imagine you have a long chain of paper clips linked together. You want to remove the last paper clip, but you only have access to the first one. You try to find the last clip by checking each one one-by-one.

The Problem

Going through each paper clip one by one to find the last one takes a lot of time and effort, especially if the chain is very long. It's easy to lose track or make mistakes, and it's slow to do every time you want to remove the last clip.

The Solution

A doubly linked list keeps track of both the next and previous paper clips. This means you can quickly jump to the last clip and remove it without checking every clip. It makes removing the last item fast and simple.

Before vs After

✗ Before

current := head
for current.next != nil {
    current = current.next
}
// remove current

✓ After

if tail != nil {
    tail = tail.prev
    if tail != nil {
        tail.next = nil
    } else {
        head = nil
    }
}

What It Enables

This lets you quickly remove the last item from a list without searching through the whole list, saving time and effort.

Real Life Example

Think of a music playlist where you want to remove the last song you added. Using a doubly linked list, you can instantly remove that last song without checking every song in the list.

Key Takeaways

Manual removal from the end is slow because you must find the last item first.

Doubly linked lists store links both ways, making end removal fast.

This operation saves time and reduces errors when managing lists.

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