Delete Node by Value in DSA Go - Time & Space Complexity

Time Complexity: Delete Node by Value

O(n)

Understanding Time Complexity

We want to understand how long it takes to remove a node by its value 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 deleteNode(head *Node, value int) *Node {
    if head == nil {
        return nil
    }
    if head.val == value {
        return head.next
    }
    current := head
    for current.next != nil {
        if current.next.val == value {
            current.next = current.next.next
            break
        }
        current = current.next
    }
    return head
}

This code removes the first node with the given value from a singly linked list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Loop through the linked list nodes one by one.
• How many times: Up to n times, where n is the number of nodes in the list.

How Execution Grows With Input

As the list grows, the time to find and delete the node grows roughly in a straight line.

Input Size (n)	Approx. Operations
10	Up to 10 checks
100	Up to 100 checks
1000	Up to 1000 checks

Pattern observation: The number of steps grows directly with the list size.

Final Time Complexity

Time Complexity: O(n)

This means the time to delete a node grows linearly with the number of nodes in the list.

Common Mistake

[X] Wrong: "Deleting a node by value always takes the same time, no matter the list size."

[OK] Correct: You must search the list to find the node, so bigger lists usually take more time.

Interview Connect

Understanding this helps you explain how linked list operations scale, a key skill in many coding discussions.

Self-Check

"What if the list was doubly linked? How would the time complexity change?"

Practice

1. What is the main purpose of the Delete Node by Value operation in a linked list?

easy

A. To find the position of a node with the given value

B. To remove the first node that contains the given value from the list

C. To add a new node with the given value at the end of the list

D. To reverse the linked list

Solution

1. Step 1: Understand the operation's goal

   The operation aims to remove a node that matches a specific value from the linked list.

2. Step 2: Identify the correct description

   Removing the first node with the given value matches To remove the first node that contains the given value from the list, while others describe different operations.

3. Final Answer:

   To remove the first node that contains the given value from the list -> Option B

4. Quick Check:

   Delete node by value means remove matching node [OK]

Hint: Delete node by value means remove matching node [OK]

Common Mistakes:

• Confusing deletion with insertion
• Thinking it finds position instead of deleting
• Assuming it reverses the list

2. Which of the following Go code snippets correctly deletes the first node with value 5 from a singly linked list starting at head?

easy

A. if head == nil { return head } if head.val == 5 { return head.next } curr := head for curr.next != nil && curr.next.val != 5 { curr = curr.next } if curr.next != nil { curr.next = curr.next.next } return head

B. if head == nil { return nil } curr := head for curr != nil && curr.val != 5 { curr = curr.next } if curr != nil { curr = curr.next } return head

C. if head.val == 5 { head = nil } curr := head for curr.next != nil { if curr.next.val == 5 { curr.next = nil } curr = curr.next } return head

D. curr := head for curr.next != nil { if curr.val == 5 { curr = curr.next.next } curr = curr.next } return head

Solution

1. Step 1: Check for empty list and head node value

   if head == nil { return head } if head.val == 5 { return head.next } curr := head for curr.next != nil && curr.next.val != 5 { curr = curr.next } if curr.next != nil { curr.next = curr.next.next } return head correctly returns head.next if head's value is 5, handling deletion at the start.

2. Step 2: Traverse and delete node with value 5

   if head == nil { return head } if head.val == 5 { return head.next } curr := head for curr.next != nil && curr.next.val != 5 { curr = curr.next } if curr.next != nil { curr.next = curr.next.next } return head uses a loop to find the node before the target and updates links properly; others miss this or incorrectly update pointers.

3. Final Answer:

   Option A code correctly deletes the first node with value 5 -> Option A

4. Quick Check:

   Proper pointer update and head check = if head == nil { return head } if head.val == 5 { return head.next } curr := head for curr.next != nil && curr.next.val != 5 { curr = curr.next } if curr.next != nil { curr.next = curr.next.next } return head [OK]

Hint: Check head value first, then update next pointers carefully [OK]

Common Mistakes:

• Not handling deletion when head node matches
• Incorrect pointer updates causing list break
• Skipping traversal or deleting wrong node

3. Given the linked list: 1 -> 3 -> 5 -> 7 -> 5 -> null, what will be the list after deleting the first node with value 5 using the standard delete-by-value function?

medium

A. 1 -> 3 -> 5 -> 7 -> null

B. 1 -> 3 -> 7 -> null

C. 3 -> 5 -> 7 -> 5 -> null

D. 1 -> 3 -> 7 -> 5 -> null

Solution

1. Step 1: Identify the first node with value 5

   The first 5 appears as the third node in the list (1 -> 3 -> 5).

2. Step 2: Remove that node and link previous to next

   Deleting this node links 3 directly to 7, resulting in 1 -> 3 -> 7 -> 5 -> null.

3. Final Answer:

   1 -> 3 -> 7 -> 5 -> null -> Option D

4. Quick Check:

   First 5 removed, rest unchanged = 1 -> 3 -> 7 -> 5 -> null [OK]

Hint: Delete only first matching node, keep rest intact [OK]

Common Mistakes:

• Deleting all nodes with value 5
• Removing wrong node (second 5)
• Breaking list links incorrectly

4. The following Go code attempts to delete a node by value from a singly linked list but causes a runtime panic. What is the error?

func deleteNode(head *Node, val int) *Node {
    curr := head
    for curr.next != nil {
        if curr.next.val == val {
            curr.next = curr.next.next
            break
        }
        curr = curr.next
    }
    return head
}

medium

A. It does not handle deletion if the head node contains the value

B. It causes infinite loop due to missing curr update

C. It tries to access curr.next.val without checking if curr.next is nil

D. It deletes all nodes with the value instead of just the first

Solution

1. Step 1: Analyze head node deletion case

   The code starts from head and checks curr.next, so if head itself has the value, it is not deleted.

2. Step 2: Confirm no panic cause

   The loop condition curr.next != nil prevents nil pointer dereference, so no panic from accessing curr.next.val.

3. Final Answer:

   It does not handle deletion if the head node contains the value -> Option A

4. Quick Check:

   Head node deletion missing = It does not handle deletion if the head node contains the value [OK]

Hint: Always check head node separately before looping [OK]

Common Mistakes:

• Ignoring head node deletion
• Assuming panic from nil pointer without checking
• Not breaking after deletion causing multiple deletes

5. You have a singly linked list with nodes containing values: 2 -> 4 -> 6 -> 4 -> 8 -> null. You want to delete all nodes with value 4 using a single pass delete-by-value function that deletes only the first matching node per call. How many times must you call this function to remove all 4s, and what will the list look like after all deletions?

hard

A. Call twice; final list: 2 -> 4 -> 6 -> 8 -> null

B. Call once; final list: 2 -> 6 -> 4 -> 8 -> null

C. Call twice; final list: 2 -> 6 -> 8 -> null

D. Call once; final list: 2 -> 6 -> 8 -> null

Solution

1. Step 1: Understand single pass deletion removes only first matching node

   Each call deletes only the first node with value 4 found.

2. Step 2: Count calls needed and final list

   First call removes the first 4 (second node), list becomes 2 -> 6 -> 4 -> 8 -> null. Second call removes the next 4 (third node), list becomes 2 -> 6 -> 8 -> null.

3. Final Answer:

   Call twice; final list: 2 -> 6 -> 8 -> null -> Option C

4. Quick Check:

   Two calls remove both 4s = Call twice; final list: 2 -> 6 -> 8 -> null [OK]

Hint: One call deletes one node; repeat until all removed [OK]

Common Mistakes:

• Assuming one call removes all matching nodes
• Miscounting nodes deleted per call
• Confusing node positions after deletions