Delete Node at End in DSA Go

Mental Model

To remove the last item from a chain, find the second last item and cut off its link to the last one.

Analogy: Imagine a line of people holding hands. To remove the last person, the second last person lets go of their hand, so the last person is no longer connected.

head -> 1 -> 2 -> 3 -> null

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> null, delete node at end

Goal: Remove the last node (3) so the list ends at 2

Step 1: start at head node (1)

head -> [curr->1] -> 2 -> 3 -> null

Why: We begin from the start to find the node before the last

Step 2: move curr to next node (2)

head -> 1 -> [curr->2] -> 3 -> null

Why: Keep moving forward to find the second last node

Step 3: curr.next points to last node (3), so set curr.next to null to remove last node

head -> 1 -> 2 -> null

Why: Cutting the link removes the last node from the list

Result:

head -> 1 -> 2 -> null

Annotated Code

DSA Go

package main

import "fmt"

type Node struct {
	value int
	next  *Node
}

// deleteAtEnd removes the last node from the linked list
func deleteAtEnd(head *Node) *Node {
	// if list is empty, return nil
	if head == nil {
		return nil
	}
	// if only one node, remove it by returning nil
	if head.next == nil {
		return nil
	}
	curr := head
	// move curr until the node before last
	for curr.next.next != nil {
		curr = curr.next
	}
	// remove last node by setting next to nil
	curr.next = nil
	return head
}

// printList prints the linked list
func printList(head *Node) {
	curr := head
	for curr != nil {
		fmt.Printf("%d", curr.value)
		if curr.next != nil {
			fmt.Printf(" -> ")
		}
		curr = curr.next
	}
	fmt.Println(" -> null")
}

func main() {
	head := &Node{value: 1}
	head.next = &Node{value: 2}
	head.next.next = &Node{value: 3}

	fmt.Print("Original list: ")
	printList(head)

	head = deleteAtEnd(head)

	fmt.Print("After deleting at end: ")
	printList(head)
}

if head == nil { return nil }

handle empty list by returning nil

if head.next == nil { return nil }

handle single node list by removing it

for curr.next.next != nil { curr = curr.next }

advance curr to second last node

curr.next = nil

remove last node by cutting link

Output

Original list: 1 -> 2 -> 3 -> null

After deleting at end: 1 -> 2 -> null

Complexity Analysis

Time: O(n) because we traverse the list once to find the second last node

Space: O(1) because we only use a few pointers, no extra space

vs Alternative: Compared to removing from the front which is O(1), removing from end requires traversal, so it is slower

Edge Cases

empty list

function returns nil without error

DSA Go

if head == nil { return nil }

single node list

function removes the only node and returns nil

DSA Go

if head.next == nil { return nil }

When to Use This Pattern

When asked to remove the last item from a linked list, look for the pattern of finding the second last node to cut its link.

Common Mistakes

Mistake: Trying to remove last node by traversing until curr.next == nil and then setting curr to nil
Fix: Instead, traverse until curr.next.next == nil and set curr.next = nil to remove last node

Summary

Removes the last node from a singly linked list by cutting the link from the second last node.

Use when you need to delete the last element in a linked list.

The key is to stop at the second last node and set its next pointer to null.

Practice

1. What does the operation Delete Node at End do in a linked list?

easy

A. Removes the last node from the linked list

B. Adds a new node at the end of the linked list

C. Removes the first node from the linked list

D. Deletes all nodes in the linked list

Solution

1. Step 1: Understand the linked list structure

   A linked list is a chain of nodes where each node points to the next node.

2. Step 2: Identify the effect of deleting the last node

   Deleting the last node means removing the node that has no next node, effectively shortening the list by one at the end.

3. Final Answer:

   Removes the last node from the linked list -> Option A

4. Quick Check:

   Delete Node at End = Remove last node [OK]

Hint: Deleting at end always removes the tail node [OK]

Common Mistakes:

• Confusing delete at end with delete at start
• Thinking it adds a node instead of removing
• Assuming it deletes all nodes

2. Which of the following Go code snippets correctly deletes the last node of a singly linked list?

easy

A. func deleteEnd(head *Node) *Node { if head == nil || head.next == nil { return nil } curr := head; for curr.next.next != nil { curr = curr.next } curr.next = nil; return head }

B. func deleteEnd(head *Node) *Node { head = head.next; return head }

C. func deleteEnd(head *Node) *Node { head = nil; return head }

D. func deleteEnd(head *Node) *Node { curr := head; for curr.next != nil { curr = curr.next } curr = nil; return head }

Solution

1. Step 1: Check for empty or single node list

   If list is empty or has one node, deleting last node results in empty list (return nil).

2. Step 2: Traverse to second last node and remove last

   Loop until curr.next.next is nil, then set curr.next to nil to remove last node.

3. Final Answer:

   Code in option A correctly deletes the last node -> Option A

4. Quick Check:

   Traverse to second last node, set next to nil [OK]

Hint: Stop at second last node, then remove last [OK]

Common Mistakes:

• Setting head to nil incorrectly
• Not stopping at second last node
• Trying to delete node by setting local pointer nil

3. What will be the output linked list after running this Go code to delete the last node?

type Node struct { val int; next *Node }
func main() {
 head := &Node{val: 1, next: &Node{val: 2, next: &Node{val: 3, next: nil}}}
 head = deleteEnd(head)
 printList(head)
}
func deleteEnd(head *Node) *Node {
 if head == nil || head.next == nil { return nil }
 curr := head
 for curr.next.next != nil { curr = curr.next }
 curr.next = nil
 return head
}
func printList(head *Node) {
 for head != nil {
  fmt.Print(head.val, " -> ")
  head = head.next
 }
 fmt.Println("nil")
}

medium

A. 2 -> 3 -> nil

B. 1 -> 2 -> 3 -> nil

C. 1 -> 2 -> nil

D. nil

Solution

1. Step 1: Understand initial list

   List is 1 -> 2 -> 3 -> nil before deletion.

2. Step 2: Delete last node

   deleteEnd removes node with value 3, so list becomes 1 -> 2 -> nil.

3. Final Answer:

   1 -> 2 -> nil -> Option C

4. Quick Check:

   Last node removed, list ends at 2 [OK]

Hint: Deleting end removes last value from printed list [OK]

Common Mistakes:

• Forgetting to update next pointer to nil
• Deleting first node instead of last
• Printing old list without deletion

4. Find the bug in this Go function that tries to delete the last node of a singly linked list:

func deleteEnd(head *Node) *Node {
 if head == nil { return nil }
 curr := head
 for curr.next != nil {
  curr = curr.next
 }
 curr = nil
 return head
}

medium

A. It does not handle empty list correctly

B. It causes a runtime panic due to nil pointer dereference

C. It deletes the entire list instead of last node

D. It does not update the second last node's next pointer

Solution

1. Step 1: Analyze the loop

   The loop moves curr to the last node (where curr.next is nil).

2. Step 2: Check deletion logic

   Setting curr = nil only changes local pointer, does not update previous node's next pointer, so last node remains linked.

3. Final Answer:

   It does not update the second last node's next pointer -> Option D

4. Quick Check:

   Must update second last node's next to nil [OK]

Hint: Update second last node's next, not local pointer [OK]

Common Mistakes:

• Setting local pointer nil instead of previous node's next
• Not handling single node list
• Assuming curr = nil deletes node

5. Given a singly linked list, how would you modify the deleteEnd function to handle the edge case when the list has only one node?

hard

A. Do nothing and return head as is

B. Return nil immediately if head.next is nil, indicating list becomes empty

C. Set head.next to nil and return head

D. Traverse the list twice to confirm single node

Solution

1. Step 1: Identify single node case

   If head.next is nil, list has only one node.

2. Step 2: Decide deletion result

   Deleting last node means list becomes empty, so return nil.

3. Final Answer:

   Return nil immediately if head.next is nil, indicating list becomes empty -> Option B

4. Quick Check:

   Single node deletion returns empty list [OK]

Hint: Check if only one node, then return nil [OK]

Common Mistakes:

• Not handling single node case causing errors
• Setting head.next to nil but returning non-empty list
• Unnecessary extra traversal