Creating a Singly Linked List from Scratch in DSA C# - Complexity Walkthrough

Time Complexity: Creating a Singly Linked List from Scratch

O(n^2)

Understanding Time Complexity

When we create a singly linked list from scratch, we add elements one by one. Understanding how the time needed grows as we add more elements helps us write efficient code.

We want to know: how does the time to build the list change as the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

public class SinglyLinkedList {
    public Node Head;

    public void Add(int value) {
        Node newNode = new Node(value);
        if (Head == null) {
            Head = newNode;
        } else {
            Node current = Head;
            while (current.Next != null) {
                current = current.Next;
            }
            current.Next = newNode;
        }
    }
}
    

This code creates a singly linked list and adds new nodes at the end one by one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop inside the Add method that moves through the list to find the last node.
• How many times: For each new node added, the loop runs once for each existing node in the list until it reaches the end.

How Execution Grows With Input

As we add more nodes, the time to add each new node grows because we must walk through the whole list to find the end.

Input Size (n)	Approx. Operations
10	About 1 + 2 + ... + 10 = 55 steps
100	About 1 + 2 + ... + 100 = 5,050 steps
1000	About 1 + 2 + ... + 1000 = 500,500 steps

Pattern observation: The total steps grow roughly like the square of the number of nodes added.

Final Time Complexity

Time Complexity: O(n2)

This means that adding n nodes one by one at the end takes time proportional to n squared, because each add walks through the list.

Common Mistake

[X] Wrong: "Adding each node takes the same constant time, so total time is O(n)."

[OK] Correct: Each add must find the end by walking through the list, so the time grows with the list length, not constant.

Interview Connect

Understanding how adding nodes affects time helps you explain trade-offs in data structure design. This skill shows you can analyze and improve code efficiency.

Self-Check

"What if we kept a reference to the last node to avoid walking the list each time? How would the time complexity change?"

Practice

1. What is the main purpose of a Singly Linked List in data structures?

easy

A. To store elements in a linear order where each element points to the next

B. To store elements in a tree-like hierarchical structure

C. To store elements with direct access by index like an array

D. To store elements in a key-value pair format

Solution

1. Step 1: Understand the structure of a singly linked list

   A singly linked list stores elements in nodes where each node points to the next node, forming a chain.

2. Step 2: Compare with other data structures

   Unlike trees or arrays, singly linked lists do not have hierarchical or direct index access but maintain a linear sequence through pointers.

3. Final Answer:

   To store elements in a linear order where each element points to the next -> Option A

4. Quick Check:

   Singly Linked List = linear chain of nodes [OK]

Hint: Remember: singly linked list nodes point only forward [OK]

Common Mistakes:

• Confusing linked lists with arrays having direct index access
• Thinking linked lists store key-value pairs
• Assuming linked lists have hierarchical structure like trees

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

easy

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

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

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

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

Solution

1. Step 1: Identify the correct node structure for singly linked list

   A node must store data and a reference to the next node, so the next pointer type should be Node.

2. Step 2: Check each option's correctness

   public class Node { public int data; public Node next; } correctly declares data as int and next as Node. Options B and C incorrectly use int for next. public class Node { public int data; public Node previous; } uses previous which is for doubly linked lists.

3. Final Answer:

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

4. Quick Check:

   Node next pointer type = Node [OK]

Hint: Next pointer must be Node type, not int [OK]

Common Mistakes:

• Using int type for next pointer instead of Node
• Confusing singly linked list with doubly linked list (previous pointer)
• Omitting access modifiers making members inaccessible

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

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

public class LinkedList {
    public Node head;
    public void Add(int data) {
        Node newNode = new Node(data);
        if (head == null) {
            head = newNode;
        } else {
            Node temp = head;
            while (temp.next != null) {
                temp = temp.next;
            }
            temp.next = newNode;
        }
    }
    public void PrintList() {
        Node temp = head;
        while (temp != null) {
            Console.Write(temp.data + " -> ");
            temp = temp.next;
        }
        Console.WriteLine("null");
    }
}

class Program {
    static void Main() {
        LinkedList list = new LinkedList();
        list.Add(10);
        list.Add(20);
        list.Add(30);
        list.PrintList();
    }
}

medium

A. null

B. 10 -> 20 -> 30 -> null

C. 10 -> 20 -> null

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

Solution

1. Step 1: Trace the Add method calls

   Adding 10 sets head to node with data 10. Adding 20 appends node with data 20 at end. Adding 30 appends node with data 30 at end.

2. Step 2: Trace the PrintList method

   PrintList starts at head and prints each node's data followed by ' -> ' until null is reached, printing "10 -> 20 -> 30 -> null".

3. Final Answer:

   10 -> 20 -> 30 -> null -> Option B

4. Quick Check:

   Nodes added in order = printed in order [OK]

Hint: Add appends at end; print shows nodes in insertion order [OK]

Common Mistakes:

• Assuming Add inserts at head instead of tail
• Forgetting to print 'null' at end
• Printing nodes in reverse order

4. Identify the error in the following C# code snippet for adding a node to a singly linked list:

public void Add(int data) {
    Node newNode = new Node(data);
    if (head == null) {
        head = newNode;
    } else {
        Node temp = head;
        while (temp != null) {
            temp = temp.next;
        }
        temp.next = newNode;
    }
}

medium

A. The while loop condition should be 'temp.next != null' instead of 'temp != null'

B. The newNode should be created outside the method

C. The head should be set to null inside the else block

D. The method should return the newNode

Solution

1. Step 1: Analyze the while loop condition

   The loop currently runs while temp != null, so it stops when temp is null, meaning temp.next is not accessible.

2. Step 2: Correct the loop to stop at last node

   The loop should run while temp.next != null to stop at the last node, so we can set temp.next = newNode safely.

3. Final Answer:

   The while loop condition should be 'temp.next != null' instead of 'temp != null' -> Option A

4. Quick Check:

   Traverse until last node (temp.next != null) [OK]

Hint: Loop until temp.next is null to find last node [OK]

Common Mistakes:

• Looping until temp is null causes null reference error
• Trying to assign next on null reference
• Misplacing node creation outside method

5. You want to create a method Reverse() in your singly linked list class that reverses the list in place. Which approach correctly reverses the list?

hard

A. Swap data values between head and tail nodes repeatedly until the middle is reached.

B. Create a new list and copy nodes from the original list in reverse order.

C. Iterate through the list, for each node set its next pointer to the previous node, updating previous and current pointers until the end.

D. Set head to null and re-add nodes in reverse order using Add method.

Solution

1. Step 1: Understand in-place reversal logic

   Reversing a singly linked list in place requires changing each node's next pointer to point to the previous node, not creating new nodes or swapping data.

2. Step 2: Evaluate options for correctness

   Iterating through the list, for each node setting its next pointer to the previous node while updating previous and current pointers until the end correctly describes the standard in-place reversal. The other options involve extra memory or inefficient operations.

3. Final Answer:

   Iterate through the list, for each node set its next pointer to the previous node, updating previous and current pointers until the end. -> Option C

4. Quick Check:

   Reverse by redirecting next pointers in place [OK]

Hint: Reverse by changing next pointers, not data or new nodes [OK]

Common Mistakes:

• Creating new lists instead of reversing in place
• Swapping data instead of pointers
• Resetting head to null losing the list