Delete Node at End in DSA Go - Time & Space Complexity

Time Complexity: Delete Node at End

O(n)

Understanding Time Complexity

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

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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
}

This code removes the last node from a singly linked list by walking to the second last node and cutting off the last.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for loop that moves through the list nodes.
• How many times: It runs once for each node until the second last node.

How Execution Grows With Input

As the list gets longer, the loop runs more times to reach the second last node.

Input Size (n)	Approx. Operations
10	About 9 steps
100	About 99 steps
1000	About 999 steps

Pattern observation: The number of steps grows roughly the same as the list size.

Final Time Complexity

Time Complexity: O(n)

This means the time to delete the last node grows linearly with the list size.

Common Mistake

[X] Wrong: "Deleting the last node is always quick because it's at the end."

[OK] Correct: We must walk through the list to find the second last node, so time grows with list length.

Interview Connect

Understanding this helps you explain how linked lists work and shows you can analyze simple loops well.

Self-Check

"What if we kept a pointer to the last node? How would the time complexity change?"

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