Why Delete from Beginning of Doubly Linked List in DSA C++?

The Big Idea

What if you could remove the first item from a chain instantly without breaking anything?

The Scenario

Imagine you have a long chain of paper clips linked together. You want to remove the first paper clip quickly without breaking the whole chain.

The Problem

If you try to remove the first paper clip by searching through the entire chain every time, it will take a lot of time and you might accidentally break the chain or lose track of the links.

The Solution

Using a doubly linked list lets you remove the first item easily by just changing a few links, without touching the rest of the chain. This keeps the chain intact and saves time.

Before vs After

✗ Before

for (Node* current = head; current != nullptr; current = current->next) {
  if (current == head) {
    // remove head manually
  }
}

✓ After

void deleteFromBeginning() {
  if (head == nullptr) return;
  Node* nodeToDelete = head;
  head = head->next;
  if (head != nullptr) head->prev = nullptr;
  delete nodeToDelete;
}

What It Enables

This operation makes it possible to efficiently manage and update sequences where the first element changes often, like task queues or undo histories.

Real Life Example

Think of a playlist where the first song finishes and is removed automatically so the next song can start playing without delay.

Key Takeaways

Manually removing the first element is slow and risky.

Doubly linked lists let you remove the first element quickly by adjusting just a few pointers.

This keeps the list intact and operations efficient.

Practice

1. What happens to the head pointer when you delete the first node from a doubly linked list?

easy

A. It points to the last node

B. It becomes null immediately

C. It moves to the second node in the list

D. It remains pointing to the deleted node

Solution

1. Step 1: Understand the head pointer role

   The head points to the first node in the doubly linked list.

2. Step 2: After deletion, update the head pointer

   When the first node is deleted, the head must point to the next node, which becomes the new first node.

3. Final Answer:

   It moves to the second node in the list -> Option C

4. Quick Check:

   Head points to new first node after deletion [OK]

Hint: Head always moves to next node after deleting first [OK]

Common Mistakes:

• Not updating head pointer after deletion
• Setting head to null without checking list size
• Leaving head pointing to deleted node

2. Which of the following is the correct syntax to delete the first node in a doubly linked list in C++?

easy

A. head = head->next; delete head->prev; head->prev = nullptr;

B. Node* temp = head; head = head->next; head->prev = nullptr; delete temp;

C. head->prev = nullptr; delete head; head = head->next;

D. delete head; head = head->next; head->prev = nullptr;

Solution

1. Step 1: Store current head in temp

   We save the current first node in a temporary pointer to delete it safely after updating head.

2. Step 2: Update head and fix previous pointer

   Set head to the next node, then set head->prev to nullptr to detach the old first node.

3. Step 3: Delete the old first node

   Finally, delete the node stored in temp to free memory.

4. Final Answer:

   Node* temp = head; head = head->next; head->prev = nullptr; delete temp; -> Option B

5. Quick Check:

   Proper order: save, update, fix pointers, delete [OK]

Hint: Save head first, update pointers, then delete saved node [OK]

Common Mistakes:

• Deleting head before updating pointers
• Not setting head->prev to nullptr
• Accessing head->prev when head is nullptr

3. What will be the output of the following code snippet after deleting the first node?

struct Node {
  int data;
  Node* prev;
  Node* next;
};

Node* head = new Node{1, nullptr, nullptr};
Node* second = new Node{2, head, nullptr};
head->next = second;

// Delete from beginning
Node* temp = head;
head = head->next;
head->prev = nullptr;
delete temp;

// Print list
Node* curr = head;
while(curr != nullptr) {
  std::cout << curr->data << " ";
  curr = curr->next;
}

medium

A. 1 2

B. No output (runtime error)

C. 1

D. 2

Solution

1. Step 1: Initial list setup

   List has two nodes: 1 (head) -> 2. Head points to node with data 1.

2. Step 2: Delete first node and update pointers

   Head moves to second node (data 2), and its prev is set to nullptr. The first node is deleted.

3. Step 3: Print remaining list

   Only node with data 2 remains, so output is "2".

4. Final Answer:

   2 -> Option D

5. Quick Check:

   After deletion, list starts at 2 [OK]

Hint: After deleting head, print starts from new head [OK]

Common Mistakes:

• Printing deleted node's data
• Not updating head->prev to nullptr
• Assuming both nodes print after deletion

4. Identify the error in the following code that deletes the first node of a doubly linked list:

void deleteBeginning(Node*& head) {
  if (head == nullptr) return;
  delete head;
  head = head->next;
  if (head != nullptr) head->prev = nullptr;
}

medium

A. Deleting head before updating head pointer causes undefined behavior

B. Missing check if head->next is nullptr

C. head->prev should be set to head->next

D. No error, code works fine

Solution

1. Step 1: Analyze deletion order

   The code deletes head first, then tries to access head->next, which is invalid after deletion.

2. Step 2: Correct order to avoid undefined behavior

   First save head->next in a temp pointer, update head, then delete old node.

3. Final Answer:

   Deleting head before updating head pointer causes undefined behavior -> Option A

4. Quick Check:

   Accessing deleted memory causes error [OK]

Hint: Always update pointers before deleting nodes [OK]

Common Mistakes:

• Deleting node before updating pointers
• Accessing deleted node's members
• Not checking for empty list

5. You have a doubly linked list with nodes: 10 <-> 20 <-> 30 <-> 40. After deleting the first node twice from the beginning, what will be the printed list?

hard

A. 30 <-> 40

B. 20 <-> 30 <-> 40

C. 10 <-> 20 <-> 30 <-> 40

D. 40

Solution

1. Step 1: Initial list nodes

   List is 10 <-> 20 <-> 30 <-> 40, head points to 10.

2. Step 2: First deletion

   Delete node 10, head moves to 20, set head->prev = nullptr. List now: 20 <-> 30 <-> 40.

3. Step 3: Second deletion

   Delete node 20, head moves to 30, set head->prev = nullptr. List now: 30 <-> 40.

4. Final Answer:

   30 <-> 40 -> Option A

5. Quick Check:

   Two deletions remove first two nodes [OK]

Hint: Each deletion moves head to next node [OK]

Common Mistakes:

• Deleting only once instead of twice
• Not updating head->prev after second deletion
• Printing old list without deletions