Delete Node by Value in DSA Go - Build from Scratch

Delete Node by Value in a Singly Linked List

📖 Scenario: Imagine you have a list of tasks to do today. Sometimes, you finish a task and want to remove it from your list. We will create a simple list of tasks and learn how to remove a task by its name.

🎯 Goal: You will build a singly linked list of tasks and write code to delete a task by its name (value). Finally, you will print the list after deletion to see the updated tasks.

📋 What You'll Learn

Create a singly linked list with exact tasks: "Laundry", "Groceries", "Homework", "Exercise"

Create a variable called taskToDelete with the value "Homework"

Write a function deleteNodeByValue that deletes the first node with the value equal to taskToDelete

Print the linked list after deletion in the format: Task1 -> Task2 -> ... -> nil

💡 Why This Matters

🌍 Real World

Linked lists are used in many applications like task scheduling, undo functionality in editors, and managing playlists where items can be added or removed dynamically.

💼 Career

Understanding linked lists and how to manipulate them is fundamental for software developers, especially when working with low-level data structures or optimizing memory usage.

1

Create the initial linked list

Create a singly linked list with nodes containing these exact tasks in order: "Laundry", "Groceries", "Homework", "Exercise". Use a struct called Node with fields task string and next *Node. Create a variable head pointing to the first node.

DSA Go

// Define Node struct and create linked list with tasks
// Your code here

Hint

Start by defining the Node struct with task and next. Then create nodes linked together with the exact tasks.

2

Add the task to delete variable

Create a variable called taskToDelete and set it to the string "Homework".

DSA Go

package main

import "fmt"

type Node struct {
	task string
	next *Node
}

func main() {
	head := &Node{task: "Laundry"}
	head.next = &Node{task: "Groceries"}
	head.next.next = &Node{task: "Homework"}
	head.next.next.next = &Node{task: "Exercise"}

// Create taskToDelete variable
// Your code here
}

Hint

Just create a string variable named taskToDelete and assign it the value "Homework".

3

Write the deleteNodeByValue function

Write a function called deleteNodeByValue that takes head *Node and value string as parameters and returns the new head *Node. It should delete the first node whose task equals value. Use this function in main to delete the node with taskToDelete.

DSA Go

package main

import "fmt"

type Node struct {
	task string
	next *Node
}

func main() {
	head := &Node{task: "Laundry"}
	head.next = &Node{task: "Groceries"}
	head.next.next = &Node{task: "Homework"}
	head.next.next.next = &Node{task: "Exercise"}

taskToDelete := "Homework"

// Call deleteNodeByValue and update head
// Your code here
}

// Write deleteNodeByValue function below
// Your code here

Hint

Check if the head node has the value to delete. If yes, return the next node as new head. Otherwise, loop through nodes to find and remove the matching node.

4

Print the linked list after deletion

Write code to print the linked list starting from head in this exact format: Task1 -> Task2 -> ... -> nil. Use a for loop with a variable current to traverse the list.

DSA Go

package main

import "fmt"

type Node struct {
	task string
	next *Node
}

func deleteNodeByValue(head *Node, value string) *Node {
	if head == nil {
		return nil
	}
	if head.task == value {
		return head.next
	}
	current := head
	for current.next != nil {
		if current.next.task == value {
			current.next = current.next.next
			break
		}
		current = current.next
	}
	return head
}

func main() {
	head := &Node{task: "Laundry"}
	head.next = &Node{task: "Groceries"}
	head.next.next = &Node{task: "Homework"}
	head.next.next.next = &Node{task: "Exercise"}

taskToDelete := "Homework"

	head = deleteNodeByValue(head, taskToDelete)

// Print the linked list in format Task1 -> Task2 -> ... -> nil
// Your code here
}

Hint

Use a for loop with a variable current starting at head. Print each task followed by ->. End with nil.

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