Delete Node at Beginning in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to delete the first node of a singly linked list.

DSA Typescript

if (this.head !== null) {
  this.head = this.head[1];
}

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

Aprev

Bvalue

Cnext

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' causes errors because singly linked lists don't have 'prev'.

Assigning head to itself does not remove the first node.

The head pointer should move to the next node to delete the first node.

2fill in blank

medium

Complete the code to check if the list is empty before deleting the first node.

DSA Typescript

if (this.head [1] null) {
  this.head = this.head.next;
}

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

A==

B!=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '!=' causes deletion attempt on empty list.

Using comparison operators like '>' or '<' is invalid for null checks.

We delete the first node only if head is not null (list is not empty).

3fill in blank

hard

Fix the error in the delete method to correctly update the head pointer.

DSA Typescript

deleteFirst() {
  if (this.head !== null) {
    this.head = this.head[1];
  }
}

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

Avalue

Bprev

Chead

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' causes runtime errors because singly linked lists don't have 'prev'.

Assigning head to itself does not remove the first node.

The head should be updated to the next node to remove the first node.

4fill in blank

hard

Fill both blanks to delete the first node and handle empty list correctly.

DSA Typescript

deleteFirst() {
  if (this.head [1] null) {
    this.head = this.head[2];
  }
}

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

A!==

B==

Cnext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using '== null' causes deletion attempt on empty list.

Using 'prev' instead of 'next' causes errors.

Check head is not null, then update head to head.next to delete first node.

5fill in blank

hard

Fill all three blanks to implement deleteFirst method that updates head and returns deleted node's value or null if empty.

DSA Typescript

deleteFirst(): number | null {
  if (this.head [1] null) {
    const deletedValue = this.head.value;
    this.head = this.head[2];
    return [3];
  }
  return null;
}

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

A!==

B==

CdeletedValue

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Returning head.value after updating head loses the deleted node's value.

Using '==' instead of '!==' causes errors on empty list.

Check head is not null, save value, update head to next node, then return saved value.

Practice

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 immediately

D. The middle node is removed

Solution

1. Step 1: Understand the head pointer role

   The head pointer points to the first node in the list.

2. Step 2: Delete the first node by moving head

   Deleting the first node means moving the head pointer to the next node.

3. Final Answer:

   The head pointer moves to the second node -> Option A

4. Quick Check:

   Delete first node = move head to next [OK]

Hint: Deleting first node means moving head to next node [OK]

Common Mistakes:

• Thinking the last node is deleted
• Assuming list becomes empty always
• Deleting a middle node instead

2. Which of the following is the correct TypeScript syntax to delete the first node in a singly linked list with head pointer head?

easy

A. head = head.next;

B. head.next = head;

C. delete head;

D. head = null.next;

Solution

1. Step 1: Understand head and next pointers

   head points to first node; head.next points to second node.

2. Step 2: Assign head to head.next to delete first node

   Setting head = head.next moves head to second node, effectively deleting first.

3. Final Answer:

   head = head.next; -> Option A

4. Quick Check:

   Correct syntax to delete first node = head = head.next; [OK]

Hint: Use head = head.next to delete first node [OK]

Common Mistakes:

• Using delete keyword incorrectly
• Assigning head.next to head
• Accessing null.next causing error

3. Given the linked list: 1 -> 2 -> 3 -> null and the code snippet:

head = head.next;

What is the resulting linked list?

medium

A. 1 -> 2 -> 3 -> null

B. 2 -> 3 -> null

C. 3 -> null

D. null

Solution

1. Step 1: Initial list and head pointer

   Head points to node with value 1, list is 1 -> 2 -> 3 -> null.

2. Step 2: Move head to head.next

   head = head.next moves head to node with value 2, removing first node.

3. Final Answer:

   2 -> 3 -> null -> Option B

4. Quick Check:

   After deletion, list starts at 2 [OK]

Hint: After head = head.next, list starts from second node [OK]

Common Mistakes:

• Assuming list remains unchanged
• Thinking last node is deleted
• Confusing node values after deletion

4. Consider this TypeScript code to delete the first node:

if (head !== null) {
  head = head.next;
}

What error does this fix compared to just head = head.next;?

medium

A. Prevents deleting the last node

B. Prevents memory leak

C. Prevents runtime error if list is empty

D. Prevents deleting middle node

Solution

1. Step 1: Understand head could be null

   If list is empty, head is null and head.next causes error.

2. Step 2: Check head before accessing next

   Condition ensures head is not null before moving to head.next, avoiding runtime error.

3. Final Answer:

   Prevents runtime error if list is empty -> Option C

4. Quick Check:

   Check head != null before head.next avoids error [OK]

Hint: Check head not null before deleting first node [OK]

Common Mistakes:

• Ignoring null head causing crash
• Thinking it prevents memory leaks
• Confusing node positions deleted

5. You have a singly linked list with nodes: 10 -> 20 -> 30 -> 40 -> null. You want to delete the first two nodes safely. Which code snippet correctly does this?

hard

A. head = null;

B. head = head.next; head = head.next;

C. head.next = head.next.next;

D. if (head && head.next) { head = head.next.next; }

Solution

1. Step 1: Understand deleting two nodes means moving head twice

   Deleting first two nodes means head should point to third node.

2. Step 2: Check head and head.next exist before moving head twice

   Condition ensures no runtime error when accessing head.next.next.

3. Final Answer:

   if (head && head.next) { head = head.next.next; } -> Option D

4. Quick Check:

   Check head and head.next before double move [OK]

Hint: Check head and head.next before skipping two nodes [OK]

Common Mistakes:

• Not checking head.next causing error
• Assigning head.next instead of head
• Setting head to null deletes entire list