Delete Node by Value in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the head node contains the value to delete.

DSA Javascript

if (this.head && this.head.value === [1]) {
  this.head = this.head.next;
  return;
}

Drag options to blanks, or click blank then click option'

Atarget

Bnode

Cval

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that is not defined like 'val' or 'target'.

Comparing to 'node' which is an object, not the value.

The value to delete is stored in the variable 'value', so we compare head.value to 'value'.

2fill in blank

medium

Complete the code to traverse the list and find the node before the one to delete.

DSA Javascript

let current = this.head;
while (current.next && current.next.value !== [1]) {
  current = current.next;
}

Drag options to blanks, or click blank then click option'

Avalue

Btarget

Cnode

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'node' or 'target' which are not the parameter name.

Comparing current.value instead of current.next.value.

We compare current.next.value to the 'value' parameter to find the node to delete.

3fill in blank

hard

Fix the error in the code to delete the node by updating the links correctly.

DSA Javascript

if (current.next) {
  current.next = [1];
}

Drag options to blanks, or click blank then click option'

Acurrent.next.next

Bcurrent.next

Ccurrent

Dthis.head

Attempts:

3 left

💡 Hint

Common Mistakes

Setting current.next to current.next (no change).

Setting current.next to current (creates a cycle).

To delete the node, we skip it by pointing current.next to current.next.next.

4fill in blank

hard

Fill both blanks to complete the delete function that removes the first node with the given value.

DSA Javascript

delete(value) {
  if (this.head && this.head.value === [1]) {
    this.head = this.head.next;
    return;
  }
  let current = this.head;
  while (current.next && current.next.value !== [2]) {
    current = current.next;
  }
  if (current.next) {
    current.next = current.next.next;
  }
}

Drag options to blanks, or click blank then click option'

Avalue

Bnode

Ctarget

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in the two comparisons.

Forgetting to check the head node separately.

Both blanks compare node values to the 'value' parameter to find and delete the node.

5fill in blank

hard

Fill all three blanks to implement a delete method that removes the first node with the given value from the linked list.

DSA Javascript

delete(value) {
  if (this.head && this.head.value === [1]) {
    this.head = this.head.next;
    return;
  }
  let current = this.head;
  while (current.next && current.next.value !== [2]) {
    current = current.next;
  }
  if (current.next) {
    current.next = [3];
  }
}

Drag options to blanks, or click blank then click option'

Avalue

Bcurrent.next.next

Cnode

Dthis.head

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names for comparison.

Setting current.next incorrectly, causing broken links.

The first two blanks compare node values to 'value'. The third blank skips the node by linking to current.next.next.

Practice

1. What is the main purpose of the deleteNodeByValue function in a linked list?

easy

A. To reverse the linked list

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

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

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

Solution

1. Step 1: Understand the function's goal

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

2. Step 2: Identify the action performed

   Deleting a node by value means finding the first node with that value and removing it from the list.

3. Final Answer:

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

4. Quick Check:

   Delete node by value = Remove node with value [OK]

Hint: Deleting by value means removing matching node, not adding [OK]

Common Mistakes:

• Confusing deletion with insertion
• Thinking it finds position only
• Assuming it reverses the list

2. Which of the following is the correct way to check if the head node contains the value to delete in JavaScript?

easy

A. if (head.value === value) { /* delete head */ }

B. if (head == value) { /* delete head */ }

C. if (head.value = value) { /* delete head */ }

D. if (head.value != value) { /* delete head */ }

Solution

1. Step 1: Understand comparison syntax in JavaScript

   Use triple equals (===) to compare values strictly without type coercion.

2. Step 2: Identify correct condition for head node value check

   Check if head.value equals the target value using ===.

3. Final Answer:

   if (head.value === value) { /* delete head */ } -> Option A

4. Quick Check:

   Use === for value comparison [OK]

Hint: Use === to compare values, not = or == [OK]

Common Mistakes:

• Using single = which assigns instead of compares
• Using == which can cause unexpected type coercion
• Comparing head node directly to value without .value

3. Consider the linked list: 10 -> 20 -> 30 -> 40 -> null. After calling deleteNodeByValue(head, 30), what will be the new list?

medium

A. 20 -> 30 -> 40 -> null

B. 10 -> 20 -> 30 -> 40 -> null

C. 10 -> 20 -> 40 -> null

D. 10 -> 30 -> 40 -> null

Solution

1. Step 1: Locate the node with value 30

   Traverse the list: 10, 20, then 30 is found as the third node.

2. Step 2: Remove the node with value 30

   Adjust the previous node's next pointer (20's next) to point to 40, skipping 30.

3. Final Answer:

   10 -> 20 -> 40 -> null -> Option C

4. Quick Check:

   Deleted 30 node, list skips it [OK]

Hint: Skip node by linking previous to next [OK]

Common Mistakes:

• Not updating previous node's next pointer
• Deleting wrong node or none
• Confusing node values or order

4. Given this code snippet to delete a node by value, what is the main bug?

function deleteNodeByValue(head, value) {
  let current = head;
  while (current.next !== null) {
    if (current.next.value === value) {
      current.next = current.next.next;
      return head;
    }
    current = current.next;
  }
  return head;
}

medium

A. It causes an infinite loop when value is not found

B. It does not handle deleting the head node if it matches the value

C. It deletes all nodes with the given value instead of just one

D. It throws an error if the list is empty

Solution

1. Step 1: Analyze head node deletion case

   The code starts checking from current.next, so it never checks if head itself has the value.

2. Step 2: Identify missing head deletion handling

   If the head node contains the value, it won't be deleted, causing incorrect behavior.

3. Final Answer:

   It does not handle deleting the head node if it matches the value -> Option B

4. Quick Check:

   Head node deletion missing = Bug [OK]

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

Common Mistakes:

• Assuming loop covers head node
• Ignoring empty list case
• Thinking it deletes all matching nodes

5. You have a singly linked list: 5 -> 10 -> 15 -> 10 -> 20 -> null. You want to delete the first node with value 10. After deletion, what is the correct list?

hard

A. 5 -> 15 -> 10 -> 20 -> null

B. 5 -> 10 -> 15 -> 20 -> null

C. 5 -> 15 -> 20 -> null

D. 10 -> 15 -> 10 -> 20 -> null

Solution

1. Step 1: Identify the first node with value 10

   Traverse from head: 5 (no), then 10 (yes) at second node.

2. Step 2: Delete the first 10 node

   Link 5's next pointer to 15, skipping the first 10 node.

3. Final Answer:

   5 -> 15 -> 10 -> 20 -> null -> Option A

4. Quick Check:

   Only first 10 removed, rest remain [OK]

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

Common Mistakes:

• Deleting all nodes with value 10
• Deleting wrong 10 node
• Breaking list links incorrectly