Create and Initialize Doubly Linked List in DSA C# - Time & Space Complexity

Time Complexity: Create and Initialize Doubly Linked List

O(n)

Understanding Time Complexity

We want to understand how long it takes to create and set up a doubly linked list when given some data.

How does the time needed change as the number of items grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public class Node {
    public int Value;
    public Node Prev;
    public Node Next;
    public Node(int value) {
        Value = value;
        Prev = null;
        Next = null;
    }
}

public Node CreateDoublyLinkedList(int[] values) {
    if (values.Length == 0) return null;
    Node head = new Node(values[0]);
    Node current = head;
    for (int i = 1; i < values.Length; i++) {
        Node newNode = new Node(values[i]);
        current.Next = newNode;
        newNode.Prev = current;
        current = newNode;
    }
    return head;
}
    

This code creates a doubly linked list from an array of integers by linking nodes forward and backward.

Identify Repeating Operations

• Primary operation: The for-loop that creates and links each new node.
• How many times: It runs once for each element after the first, so roughly n-1 times for n elements.

How Execution Grows With Input

Each new item requires creating a node and linking it, so the work grows steadily as the list grows.

Input Size (n)	Approx. Operations
10	About 9 node creations and links
100	About 99 node creations and links
1000	About 999 node creations and links

Pattern observation: The number of operations grows directly with the number of items.

Final Time Complexity

Time Complexity: O(n)

This means the time to create the list grows in a straight line with the number of elements.

Common Mistake

[X] Wrong: "Creating a doubly linked list takes constant time because each node is created once."

[OK] Correct: Even though each node is created once, the total time depends on how many nodes there are, so it grows with the list size.

Interview Connect

Understanding how list creation time grows helps you explain and reason about data structure setup in interviews confidently.

Self-Check

"What if we changed the input from an array to a linked list? How would the time complexity change?"

Practice

1. What is the main purpose of a doubly linked list in data structures?

easy

A. To allow traversal in both forward and backward directions

B. To store data in a fixed-size array

C. To sort data automatically

D. To store data only at the beginning

Solution

1. Step 1: Understand doubly linked list structure

   A doubly linked list has nodes with pointers to both next and previous nodes.

2. Step 2: Identify traversal capability

   This structure allows moving forward and backward through the list easily.

3. Final Answer:

   To allow traversal in both forward and backward directions -> Option A

4. Quick Check:

   Doubly linked list = bidirectional traversal [OK]

Hint: Remember: doubly means two directions [OK]

Common Mistakes:

• Confusing doubly linked list with arrays
• Thinking it sorts data automatically
• Assuming it only allows forward traversal

2. Which of the following is the correct way to declare a node class for a doubly linked list in C#?

easy

A. class Node { public int data; public Node next; }

B. public class Node { int data; int next; int prev; }

C. public class Node { public int data; public Node next; public Node prev; }

D. public class Node { public int data; public Node next; private Node previous; }

Solution

1. Step 1: Check node fields for doubly linked list

   Node must have data, next pointer, and previous pointer of type Node.

2. Step 2: Verify correct syntax and access modifiers

   public class Node { public int data; public Node next; public Node prev; } declares public int data, and public Node next and prev correctly.

3. Final Answer:

   public class Node { public int data; public Node next; public Node prev; } -> Option C

4. Quick Check:

   Node class with data, next, prev = public class Node { public int data; public Node next; public Node prev; } [OK]

Hint: Node needs next and prev as Node types [OK]

Common Mistakes:

• Using int for next and prev instead of Node
• Missing prev pointer for doubly linked list
• Incorrect access modifiers

3. What will be the output of the following C# code that creates a doubly linked list with nodes 10, 20, 30 and prints forward traversal?

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

public class DoublyLinkedList {
    public Node head;
    public void AddLast(int data) {
        Node newNode = new Node(data);
        if (head == null) {
            head = newNode;
            return;
        }
        Node temp = head;
        while (temp.next != null) temp = temp.next;
        temp.next = newNode;
        newNode.prev = temp;
    }
    public void PrintForward() {
        Node temp = head;
        while (temp != null) {
            Console.Write(temp.data + " -> ");
            temp = temp.next;
        }
        Console.WriteLine("null");
    }
}

class Program {
    static void Main() {
        DoublyLinkedList list = new DoublyLinkedList();
        list.AddLast(10);
        list.AddLast(20);
        list.AddLast(30);
        list.PrintForward();
    }
}

medium

A. 30 -> 20 -> 10 -> null

B. null

C. 10 -> 20 -> null

D. 10 -> 20 -> 30 -> null

Solution

1. Step 1: Understand AddLast method behavior

   AddLast adds nodes at the end, so nodes are 10, then 20, then 30 in order.

2. Step 2: Trace PrintForward output

   PrintForward prints from head to last node: 10 -> 20 -> 30 -> null.

3. Final Answer:

   10 -> 20 -> 30 -> null -> Option D

4. Quick Check:

   Forward print = 10 -> 20 -> 30 -> null [OK]

Hint: AddLast appends; PrintForward prints from head [OK]

Common Mistakes:

• Confusing forward and backward traversal
• Missing last node in output
• Printing in reverse order

4. Identify the error in the following code snippet that tries to add a node at the end of a doubly linked list:

public void AddLast(int data) {
    Node newNode = new Node(data);
    if (head == null) {
        head = newNode;
        return;
    }
    Node temp = head;
    while (temp.next != null) {
        temp = temp.next;
    }
    temp.next = newNode;
    // Missing line here
}

medium

A. Missing setting newNode.prev = temp

B. Missing setting head = newNode

C. Missing while loop to find last node

D. Missing creating newNode instance

Solution

1. Step 1: Review AddLast logic

   After finding last node, newNode is linked as next of temp.

2. Step 2: Check doubly linked list pointer updates

   For doubly linked list, newNode.prev must point back to temp to maintain backward link.

3. Final Answer:

   Missing setting newNode.prev = temp -> Option A

4. Quick Check:

   Set newNode.prev to temp for backward link [OK]

Hint: Always set newNode.prev when adding at end [OK]

Common Mistakes:

• Forgetting to set newNode.prev
• Reassigning head incorrectly
• Skipping loop to find last node

5. You want to create a doubly linked list from an array of integers {1, 2, 3, 4} such that the list is 1 -> 2 -> 3 -> 4 (head points to 1), and then reverse print the list (4 -> 3 -> 2 -> 1 -> null) using backward traversal. Which approach correctly initializes the list and prints it backward?

int[] arr = {1, 2, 3, 4};
DoublyLinkedList list = new DoublyLinkedList();
// Step 1: Add all elements from arr to list
// Step 2: Print list backward

// Which code snippet correctly does this?

hard

A. for (int i = arr.Length - 1; i >= 0; i--) list.AddLast(arr[i]);
Node temp = list.head;
while (temp != null) { Console.Write(temp.data + " -> "); temp = temp.next; }
Console.WriteLine("null");

B. foreach (int val in arr) list.AddLast(val);
Node temp = list.head;
while (temp.next != null) temp = temp.next;
while (temp != null) { Console.Write(temp.data + " -> "); temp = temp.prev; }
Console.WriteLine("null");

C. foreach (int val in arr) list.AddFirst(val);
Node temp = list.head;
while (temp != null) { Console.Write(temp.data + " -> "); temp = temp.next; }
Console.WriteLine("null");

D. for (int i = 0; i < arr.Length; i++) list.AddLast(arr[i]);
Node temp = list.head;
while (temp != null) { Console.Write(temp.data + " -> "); temp = temp.next; }
Console.WriteLine("null");

Solution

1. Step 1: Add elements in order using AddLast

   Use foreach (int val in arr) list.AddLast(val); to build the list as 1 -> 2 -> 3 -> 4, preserving array order.

2. Step 2: Traverse to last node and print backward

   Node temp = list.head; while (temp.next != null) temp = temp.next; then while (temp != null) { Console.Write(temp.data + " -> "); temp = temp.prev; } Console.WriteLine("null"); to print 4 -> 3 -> 2 -> 1 -> null.

3. Final Answer:

   foreach (int val in arr) list.AddLast(val); Node temp = list.head; while (temp.next != null) temp = temp.next; while (temp != null) { Console.Write(temp.data + " -> "); temp = temp.prev; } Console.WriteLine("null"); -> Option B

4. Quick Check:

   Build forward list + backward traversal from tail [OK]

Hint: Add in order, then move to tail to print backward [OK]

Common Mistakes:

• Adding elements in reverse order
• Printing forward instead of backward
• Using AddFirst which reverses order