Delete Node at End in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Go

for current := head; current != nil && current.[1] != nil; current = current.next {
    // loop to find last node
}

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

Aprev

Bhead

Cvalue

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' to move forward.

Trying to access 'value' to move to next node.

We use current.next to move to the next node until we reach the last node where next is nil.

2fill in blank

medium

Complete the code to update the second last node's next pointer to nil, effectively deleting the last node.

DSA Go

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

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

Anext

Bprev

Cvalue

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Setting 'prev' to nil instead of 'next'.

Trying to set 'value' to nil which is invalid.

We check if current.next is not nil, then set current.next to nil to remove the last node.

3fill in blank

hard

Fix the error in the code to handle the case when the list has only one node.

DSA Go

if head != nil && head.[1] == nil {
    head = nil
}

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

Aprev

Bnext

Cvalue

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Checking head.prev which does not exist in singly linked list.

Checking head.value instead of head.next.

If head.next is nil, it means there is only one node, so setting head to nil deletes it.

4fill in blank

hard

Fill both blanks to traverse the list and keep track of the previous node.

DSA Go

prev := head
current := head.[1]
for current != nil && current.[2] != nil {
    prev = current
    current = current.next
}

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

Anext

Bprev

Cvalue

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' to move current forward.

Using 'value' instead of 'next' to traverse.

We start current at head.next and move forward using current.next while updating prev.

5fill in blank

hard

Fill all three blanks to delete the last node and return the updated head.

DSA Go

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

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

Anext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' in conditions.

Not handling the case when list has only one node.

We check if head.next is nil, start current at head.next, and loop while current.next is not nil to find last node.

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