Delete Node at Beginning in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output after deleting the first node in a singly linked list

What is the printed state of the linked list after deleting the first node?

DSA Typescript

class Node {
  data: number;
  next: Node | null;
  constructor(data: number) {
    this.data = data;
    this.next = null;
  }
}

class LinkedList {
  head: Node | null = null;

  insertAtEnd(data: number) {
    const newNode = new Node(data);
    if (!this.head) {
      this.head = newNode;
      return;
    }
    let current = this.head;
    while (current.next) {
      current = current.next;
    }
    current.next = newNode;
  }

  deleteAtBeginning() {
    if (!this.head) return;
    this.head = this.head.next;
  }

  printList() {
    let current = this.head;
    let result = '';
    while (current) {
      result += current.data + ' -> ';
      current = current.next;
    }
    result += 'null';
    console.log(result);
  }
}

const list = new LinkedList();
list.insertAtEnd(10);
list.insertAtEnd(20);
list.insertAtEnd(30);
list.deleteAtBeginning();
list.printList();

A 20 -> 30 -> null

B 10 -> 20 -> 30 -> null

C 30 -> null

D null

💡 Hint

Deleting the first node means the head moves to the next node.

Initially, the list is 10 -> 20 -> 30 -> null. After deleting the first node (10), the head points to 20, so the list becomes 20 -> 30 -> null.

❓ Predict Output

intermediate

Output when deleting from an empty linked list

What is the output when deleteAtBeginning is called on an empty linked list?

DSA Typescript

class Node {
  data: number;
  next: Node | null;
  constructor(data: number) {
    this.data = data;
    this.next = null;
  }
}

class LinkedList {
  head: Node | null = null;

  deleteAtBeginning() {
    if (!this.head) return;
    this.head = this.head.next;
  }

  printList() {
    let current = this.head;
    let result = '';
    while (current) {
      result += current.data + ' -> ';
      current = current.next;
    }
    result += 'null';
    console.log(result);
  }
}

const list = new LinkedList();
list.deleteAtBeginning();
list.printList();

A Empty list

B null

C Error: Cannot read property 'next' of null

D undefined

💡 Hint

Deleting from empty list should not change anything.

Since the list is empty (head is null), deleteAtBeginning does nothing. Printing the list shows 'null'.

🔧 Debug

advanced

Identify the error in deleteAtBeginning method

What error will occur when running this code and why?

DSA Typescript

class Node {
  data: number;
  next: Node | null;
  constructor(data: number) {
    this.data = data;
    this.next = null;
  }
}

class LinkedList {
  head: Node | null = null;

  deleteAtBeginning() {
    this.head = this.head.next;
  }
}

const list = new LinkedList();
list.deleteAtBeginning();

A No error, deletes first node

B SyntaxError: Unexpected token

C ReferenceError: head is not defined

D TypeError: Cannot read property 'next' of null

💡 Hint

Check what happens if head is null before accessing next.

head is null initially, so accessing head.next causes a TypeError.

❓ Predict Output

advanced

Output after multiple deletions at beginning

What is the printed state of the linked list after deleting the first node twice?

DSA Typescript

class Node {
  data: number;
  next: Node | null;
  constructor(data: number) {
    this.data = data;
    this.next = null;
  }
}

class LinkedList {
  head: Node | null = null;

  insertAtEnd(data: number) {
    const newNode = new Node(data);
    if (!this.head) {
      this.head = newNode;
      return;
    }
    let current = this.head;
    while (current.next) {
      current = current.next;
    }
    current.next = newNode;
  }

  deleteAtBeginning() {
    if (!this.head) return;
    this.head = this.head.next;
  }

  printList() {
    let current = this.head;
    let result = '';
    while (current) {
      result += current.data + ' -> ';
      current = current.next;
    }
    result += 'null';
    console.log(result);
  }
}

const list = new LinkedList();
list.insertAtEnd(5);
list.insertAtEnd(15);
list.insertAtEnd(25);
list.deleteAtBeginning();
list.deleteAtBeginning();
list.printList();

A 5 -> 15 -> 25 -> null

B 15 -> 25 -> null

C 25 -> null

D null

💡 Hint

Each deletion removes the current head node.

After first deletion, list is 15 -> 25 -> null. After second deletion, list is 25 -> null.

🧠 Conceptual

expert

Effect of deleting the first node on list length

If a singly linked list has 7 nodes, what will be the length after deleting the first node?

A 6

B 7

C 5

D 0

💡 Hint

Deleting the first node removes exactly one node from the list.

Deleting the first node reduces the list length by one, so from 7 to 6.

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