Delete Node by Value in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal when deleting a node by value in a linked list?

To find the node with the given value and remove it from the list while keeping the list connected.

beginner

In a singly linked list, what special case must you handle when deleting a node by value?

Deleting the head node requires updating the head pointer to the next node.

beginner

Why do we keep track of the previous node when deleting a node by value in a singly linked list?

Because we need to update the previous node's next pointer to skip the deleted node and keep the list connected.

beginner

What happens if the value to delete is not found in the linked list?

The list remains unchanged because no node matches the value to delete.

intermediate

Show the Go code snippet to delete a node by value from a singly linked list.

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

What should you do if the node to delete is the head of the linked list?

A Update the head pointer to the next node

B Delete the entire list

C Do nothing

D Delete the last node instead

When deleting the head node, update the head pointer to the next node to keep the list connected.

Why do we need to keep track of the previous node when deleting a node in a singly linked list?

A To delete the previous node as well

B To update the previous node's next pointer to skip the deleted node

C To find the last node

D To reverse the list

The previous node's next pointer must be updated to skip the node being deleted.

What happens if the value to delete is not found in the list?

A The list remains unchanged

B The list is cleared

C The last node is deleted

D An error is thrown

If the value is not found, no changes are made to the list.

Which of these is the correct way to delete a node by value in Go?

A Only delete if node is at the end

B Delete node without updating pointers

C Traverse list, update previous node's next pointer when value found

D Create a new list without the value

You must traverse and update pointers to remove the node properly.

If the linked list is empty, what should the delete function return?

A An error

B A new node

C The original list

D nil

If the list is empty (head is nil), return nil as there is nothing to delete.

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

Think about the head node and how to keep the list connected.

Write a Go function to delete a node by value from a singly linked list and explain how it works.

Focus on pointer updates and edge cases.

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