Insert at Beginning of Doubly Linked List in DSA Java - Time & Space Complexity

Time Complexity: Insert at Beginning of Doubly Linked List

O(1)

Understanding Time Complexity

We want to understand how the time needed to insert a new node at the start of a doubly linked list changes as the list grows.

Specifically, how does the number of steps change when the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class Node {
  int data;
  Node prev, next;
  Node(int d) { data = d; prev = next = null; }
}

Node insertAtBeginning(Node head, int data) {
  Node newNode = new Node(data);
  newNode.next = head;
  if (head != null) head.prev = newNode;
  newNode.prev = null;
  return newNode;
}
    

This code creates a new node and inserts it at the start of the doubly linked list.

Identify Repeating Operations

Look for loops or repeated steps that depend on list size.

• Primary operation: Creating and linking one new node.
• How many times: Only once per insertion, no loops or traversals.

How Execution Grows With Input

Adding a node at the start always takes the same number of steps, no matter how long the list is.

Input Size (n)	Approx. Operations
10	5 steps (create node, link pointers)
100	5 steps (same as above)
1000	5 steps (still the same)

Pattern observation: The time stays constant regardless of list size.

Final Time Complexity

Time Complexity: O(1)

This means inserting at the beginning takes a fixed amount of time, no matter how big the list is.

Common Mistake

[X] Wrong: "Inserting at the start takes longer as the list grows because we have to move all nodes."

[OK] Correct: We only change a few pointers at the start; we never move or touch the other nodes.

Interview Connect

Knowing this helps you explain why linked lists are good for quick insertions at the start, a common question in interviews.

Self-Check

"What if we had to insert at the end of the doubly linked list without a tail pointer? How would the time complexity change?"

Practice

1. What is the main purpose of inserting a node at the beginning of a doubly linked list?

easy

A. To add a new node after the tail node

B. To reverse the entire list

C. To delete the first node of the list

D. To add a new node before the current head node

Solution

1. Step 1: Understand the position of insertion

   Inserting at the beginning means placing the new node before the current head node.

2. Step 2: Identify the effect on the list

   The new node becomes the new head, and its next pointer points to the old head.

3. Final Answer:

   To add a new node before the current head node -> Option D

4. Quick Check:

   Insert at beginning = add before head [OK]

Hint: Beginning insertion means new node becomes the head [OK]

Common Mistakes:

• Confusing beginning with end insertion
• Thinking insertion deletes nodes
• Mixing up singly and doubly linked list behavior

2. Which of the following is the correct syntax to create a new node in Java for a doubly linked list?

easy

A. Node newNode = Node(data); newNode.prev = null; newNode.next = null;

B. Node newNode = new Node(data); newNode.prev = null; newNode.next = null;

C. Node newNode = new Node(); newNode.data = data; newNode.prev = null; newNode.next = null;

D. Node newNode = new Node(data); newNode.prev = data; newNode.next = data;

Solution

1. Step 1: Understand Java object creation syntax

   In Java, objects are created using the 'new' keyword followed by the constructor.

2. Step 2: Check node initialization

   The node's previous and next pointers should be set to null initially.

3. Final Answer:

   Node newNode = new Node(data); newNode.prev = null; newNode.next = null; -> Option B

4. Quick Check:

   Java new object = new Constructor() [OK]

Hint: Use 'new' keyword and constructor to create objects [OK]

Common Mistakes:

• Omitting 'new' keyword
• Assigning data to prev or next pointers
• Using incorrect constructor syntax

3. What will be the output of the doubly linked list after inserting 10 at the beginning of the list containing nodes 20 <-> 30?

medium

A. 10 <-> 20 <-> 30

B. 20 <-> 30 <-> 10

C. 30 <-> 20 <-> 10

D. 10 <-> 30 <-> 20

Solution

1. Step 1: Identify current list order

   The list currently is 20 <-> 30, with 20 as head and 30 as tail.

2. Step 2: Insert 10 at beginning

   New node 10 becomes head, its next points to 20, and 20's prev points to 10.

3. Final Answer:

   10 <-> 20 <-> 30 -> Option A

4. Quick Check:

   Insert at beginning adds node before head [OK]

Hint: New node at start shifts old head right [OK]

Common Mistakes:

• Appending instead of prepending
• Not updating previous pointer of old head
• Reversing node order incorrectly

4. Given the following code snippet for inserting at the beginning of a doubly linked list, what is the error?

void insertAtBeginning(int data) {
  Node newNode = new Node(data);
  newNode.next = head;
  head.prev = newNode;
  head = newNode;
}

medium

A. No error, code works correctly

B. head should be updated before setting newNode.next

C. head.prev = newNode; causes NullPointerException if list is empty

D. newNode.next should be null

Solution

1. Step 1: Analyze code for empty list case

   If the list is empty, head is null, so head.prev causes NullPointerException.

2. Step 2: Identify fix

   Check if head is not null before setting head.prev = newNode.

3. Final Answer:

   head.prev = newNode; causes NullPointerException if list is empty -> Option C

4. Quick Check:

   Null head means head.prev causes error [OK]

Hint: Check for null head before accessing head.prev [OK]

Common Mistakes:

• Ignoring empty list case
• Assuming head is never null
• Misordering pointer updates

5. You have a doubly linked list with nodes: 5 <-> 15 <-> 25. You want to insert a new node with value 1 at the beginning and then reverse the list. What will be the final list after these operations?

hard

A. 25 <-> 15 <-> 5 <-> 1

B. 1 <-> 5 <-> 15 <-> 25

C. 1 <-> 25 <-> 15 <-> 5

D. 5 <-> 15 <-> 25 <-> 1

Solution

1. Step 1: Insert 1 at beginning

   List becomes 1 <-> 5 <-> 15 <-> 25 after insertion.

2. Step 2: Reverse the list

   Reversing swaps head and tail, so list becomes 25 <-> 15 <-> 5 <-> 1.

3. Final Answer:

   25 <-> 15 <-> 5 <-> 1 -> Option A

4. Quick Check:

   Insert at start then reverse flips order [OK]

Hint: Insert first, then reverse flips entire list [OK]

Common Mistakes:

• Reversing before insertion
• Not updating prev pointers when reversing
• Confusing node order after reverse