Delete Node at End in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does 'Delete Node at End' mean in a linked list?

It means removing the last node from a linked list, so the second last node becomes the new end.

beginner

In a singly linked list, why do we need to traverse the list to delete the last node?

Because we only have a pointer to the head and each node points forward, we must find the second last node to update its next pointer to nil.

beginner

What happens if you delete the last node in a linked list with only one node?

The list becomes empty, so the head pointer is set to nil.

intermediate

Show the Go code snippet to delete the last node in a singly linked list.

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

beginner

Why is it important to handle the case when the list is empty before deleting the last node?

Because trying to delete from an empty list can cause errors or crashes since there is no node to delete.

What pointer do you update when deleting the last node in a singly linked list?

A The next pointer of the last node

B The head pointer

C The next pointer of the second last node

D The previous pointer of the last node

You update the next pointer of the second last node to nil to remove the last node.

What should you do if the linked list has only one node and you delete the last node?

A Set head to nil

B Set head to the same node

C Do nothing

D Set next pointer of head to itself

Deleting the only node means the list becomes empty, so head should be set to nil.

Why do you need to traverse the list to delete the last node in a singly linked list?

A Because you need to find the second last node to update its next pointer

B Because the last node has no next pointer

C Because the head pointer points to the last node

D Because you need to delete all nodes before the last

You must find the second last node to set its next pointer to nil, removing the last node.

What happens if you try to delete the last node from an empty linked list?

A The list crashes

B Nothing happens, list remains empty

C The head pointer is set to a new node

D The list becomes circular

Deleting from an empty list does nothing; the list stays empty.

In Go, what is the return type of a function that deletes the last node from a linked list?

A string

B int

C bool

D *Node (pointer to head node)

The function returns the updated head pointer of type *Node.

Explain step-by-step how to delete the last node from a singly linked list.

Think about edge cases and pointer updates.

Write a Go function to delete the last node of a singly linked list and explain how it works.

Focus on pointer manipulation and edge cases.

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