Delete from End of Doubly Linked List in DSA Go - Time & Space Complexity

Time Complexity: Delete from End of Doubly Linked List

O(1)

Understanding Time Complexity

We want to understand how long it takes to remove the last item from a doubly linked list.

How does the time needed change as the list grows bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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
}

This code removes the last node from a doubly linked list by updating pointers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Accessing the tail node and updating pointers.
• How many times: This happens only once per call, no loops or recursion.

How Execution Grows With Input

Removing the last node takes the same steps no matter how big the list is.

Input Size (n)	Approx. Operations
10	5-6 steps
100	5-6 steps
1000	5-6 steps

Pattern observation: The number of steps stays about the same regardless of list size.

Final Time Complexity

Time Complexity: O(1)

This means deleting from the end takes a constant amount of time no matter how large the list is.

Common Mistake

[X] Wrong: "Deleting from the end takes longer as the list grows because you have to find the last node."

[OK] Correct: The doubly linked list keeps a direct pointer to the last node (tail), so no searching is needed.

Interview Connect

Knowing that deleting from the end is quick helps you choose the right data structure for fast removals.

Self-Check

"What if we did not keep a tail pointer? How would the time complexity change?"

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