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DSA Goprogramming~5 mins

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

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Time Complexity: Delete Node at Beginning
O(1)
Understanding Time Complexity

We want to know how the time to delete the first node in a linked list changes as the list grows.

How much work does it take to remove the first item, no matter the list size?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


// Delete the first node from a singly linked list
func deleteAtBeginning(head *Node) *Node {
    if head == nil {
        return nil
    }
    return head.Next
}
    

This code removes the first node by returning the second node as the new head.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: Accessing the first node and its next pointer once.
  • How many times: Exactly one time, no loops or recursion.
How Execution Grows With Input

Deleting the first node always takes the same steps, no matter how big the list is.

Input Size (n)Approx. Operations
101
1001
10001

Pattern observation: The work stays constant as the list grows.

Final Time Complexity

Time Complexity: O(1)

This means deleting the first node takes the same amount of time no matter how many nodes are in the list.

Common Mistake

[X] Wrong: "Deleting the first node takes longer if the list is bigger because we have to check all nodes."

[OK] Correct: We only change the head pointer to the next node, so we do not look at other nodes at all.

Interview Connect

Understanding this simple operation helps build confidence with linked lists and shows you can reason about efficient updates.

Self-Check

"What if we had to delete the last node instead? How would the time complexity change?"

Practice

(1/5)
1. What happens when you delete the first node in a singly linked list?
easy
A. The head pointer moves to the second node
B. The last node is deleted
C. The list becomes empty regardless of nodes
D. The middle node is deleted

Solution

  1. Step 1: Understand the head pointer role

    The head pointer points to the first node in the list.
  2. Step 2: Effect of deleting the first node

    Deleting the first node means moving the head pointer to the next node, which becomes the new first node.
  3. Final Answer:

    The head pointer moves to the second node -> Option A
  4. Quick Check:

    Deleting first node updates head = next node [OK]
Hint: Deleting first node means head points to next node [OK]
Common Mistakes:
  • Thinking last node is deleted
  • Assuming list becomes empty always
  • Deleting middle node by mistake
2. Which of the following Go code snippets correctly deletes the first node of a singly linked list with head pointer head?
easy
A. head.next = nil
B. head.next = head
C. head = nil.next
D. head = head.next

Solution

  1. Step 1: Understand pointer update to delete first node

    To delete the first node, head must point to the second node, which is head.next.
  2. Step 2: Check each option for correctness

    head = head.next correctly assigns head = head.next. Others either assign incorrectly or cause errors.
  3. Final Answer:

    head = head.next -> Option D
  4. Quick Check:

    head = head.next deletes first node [OK]
Hint: Assign head to head.next to delete first node [OK]
Common Mistakes:
  • Assigning head.next to head (wrong direction)
  • Using nil.next which causes error
  • Setting head.next to nil deletes wrong node
3. Given the following Go code for a singly linked list, what will be printed after deleting the first node?
package main
import "fmt"

type Node struct {
    value int
    next  *Node
}

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

    // Delete first node
    head = head.next

    // Print list
    current := head
    for current != nil {
        fmt.Print(current.value, " -> ")
        current = current.next
    }
    fmt.Print("null")
}
medium
A. 2 -> 3 -> null
B. 1 -> 2 -> 3 -> null
C. 3 -> null
D. null

Solution

  1. Step 1: Initial list nodes and values

    The list starts as 1 -> 2 -> 3 -> null.
  2. Step 2: Deleting first node by head = head.next

    Head moves from node with value 1 to node with value 2, so list becomes 2 -> 3 -> null.
  3. Final Answer:

    2 -> 3 -> null -> Option A
  4. Quick Check:

    After deletion, list starts at second node [OK]
Hint: After deletion, list starts from second node [OK]
Common Mistakes:
  • Printing original list without deletion
  • Assuming first node still prints
  • Confusing null with empty list
4. The following Go code attempts to delete the first node of a singly linked list but causes a runtime error. What is the error and how to fix it?
package main

type Node struct {
    value int
    next  *Node
}

func main() {
    var head *Node = nil

    // Delete first node
    head = head.next
}
medium
A. Compile error: cannot assign head.next to head
B. Runtime panic: nil pointer dereference; fix by checking if head is nil before deletion
C. No error; code works fine
D. Syntax error: missing import; fix by adding fmt package

Solution

  1. Step 1: Identify cause of runtime panic

    head is nil, so accessing head.next causes nil pointer dereference panic.
  2. Step 2: Fix by checking if head is nil before deleting

    Check if head != nil before assigning head = head.next to avoid panic.
  3. Final Answer:

    Runtime panic: nil pointer dereference; fix by checking if head is nil before deletion -> Option B
  4. Quick Check:

    Nil check prevents runtime panic [OK]
Hint: Always check head != nil before deleting first node [OK]
Common Mistakes:
  • Ignoring nil check causing panic
  • Assuming head is never nil
  • Confusing syntax and runtime errors
5. You have a singly linked list with nodes: 10 -> 20 -> 30 -> 40 -> null. Write a Go function to delete the first node and return the new head. What will be the list after deletion?
func deleteFirstNode(head *Node) *Node {
    if head == nil {
        return nil
    }
    head = head.next
    return head
}
hard
A. 30 -> 40 -> null
B. 10 -> 20 -> 30 -> 40 -> null
C. 20 -> 30 -> 40 -> null
D. null

Solution

  1. Step 1: Understand function logic

    If head is nil, return nil. Otherwise, move head to head.next and return it.
  2. Step 2: Apply function to list 10 -> 20 -> 30 -> 40 -> null

    Deleting first node (10) moves head to node with value 20, so list becomes 20 -> 30 -> 40 -> null.
  3. Final Answer:

    20 -> 30 -> 40 -> null -> Option C
  4. Quick Check:

    Function returns list starting from second node [OK]
Hint: Return head.next to delete first node and update list [OK]
Common Mistakes:
  • Returning original head without deletion
  • Not handling nil head case
  • Deleting wrong node by mistake
Recommended

Ready to prove it?

Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

10 questions~5 mineasy hard