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

Time Complexity: Creating a Singly Linked List from Scratch

O(n²)

Understanding Time Complexity

When we create a singly linked list from scratch, we add nodes one by one. Understanding how long this takes helps us know how the process grows as the list gets bigger.

We want to answer: How does the time to build the list change as we add more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class Node {
    int data;
    Node next;
    Node(int data) { this.data = data; this.next = null; }
}

class SinglyLinkedList {
    Node head;
    void add(int data) {
        Node newNode = new Node(data);
        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 that moves through the list to find the last node.
• How many times: For each add, it runs once for each node already in the list, so it grows as the list grows.

How Execution Grows With Input

Each time we add a node, we walk through the existing list to find the end. So adding the first node is fast, but adding the 10th node means walking through 9 nodes first.

Input Size (n)	Approx. Operations
10	0 + 1 + ... + 9 = 45 steps
100	0 + 1 + ... + 99 = 4950 steps
1000	0 + 1 + ... + 999 = 499,500 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 adding n nodes one by one at the end takes time that grows roughly like n times n, because each add walks through the list so far.

Common Mistake

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

[OK] Correct: Each add after the first must walk through the existing list, so the time grows with the list size, not constant.

Interview Connect

Understanding this helps you explain why adding at the end of a singly linked list without extra pointers is slower as the list grows. It shows you can spot where time is spent and how to improve it.

Self-Check

"What if we kept a pointer to the last node and updated it on each add? 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 sequence where each element points to the next

B. To store elements with keys and values for fast lookup

C. To store elements in a sorted array for quick access

D. To store elements in a tree structure with parent and child nodes

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 arrays or trees, singly linked lists do not store keys or have parent-child relationships; they only link sequentially.

3. Final Answer:

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

4. Quick Check:

   Singly Linked List = sequence with next pointers [OK]

Hint: Remember: singly linked list nodes link one after another [OK]

Common Mistakes:

• Confusing linked list with arrays or hash maps
• Thinking linked list stores key-value pairs
• Assuming linked list nodes have multiple pointers

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

easy

A. class Node { int data; Node left; Node right; }

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

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

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

Solution

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

   A singly linked list node contains data and a reference to the next node only.

2. Step 2: Check each option

   class Node { int data; Node next; } has data and next pointer; class Node { int data; Node prev; } has prev (used in doubly linked list); class Node { int data; Node left; Node right; } has left and right (used in trees); class Node { int data; int next; } incorrectly uses int for next pointer.

3. Final Answer:

   class Node { int data; Node next; } -> Option B

4. Quick Check:

   Node = data + next pointer [OK]

Hint: Node must have 'next' pointer, not 'prev' or others [OK]

Common Mistakes:

• Using 'prev' pointer instead of 'next'
• Using primitive type for next pointer
• Confusing tree node with linked list node

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

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

public class LinkedList {
  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) {
      System.out.print(temp.data + " -> ");
      temp = temp.next;
    }
    System.out.print("null");
  }

  public static void main(String[] args) {
    LinkedList list = new LinkedList();
    list.add(10);
    list.add(20);
    list.add(30);
    list.printList();
  }
}

medium

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

B. null

C. 10 -> 20 -> null

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

Solution

1. Step 1: Understand the add method behavior

   The add method appends new nodes at the end of the list, preserving insertion order.

2. Step 2: Trace the printList method

   It prints nodes from head to last, showing 10, then 20, then 30, followed by null.

3. Final Answer:

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

4. Quick Check:

   Adding nodes appends at end, print shows 10->20->30->null [OK]

Hint: Add appends at end; print shows order added [OK]

Common Mistakes:

• Assuming add inserts at head instead of tail
• Printing nodes in reverse order
• Missing the final 'null' in output

4. Identify the error in the following code that attempts to add a node to the singly linked list:

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

public class LinkedList {
  Node head;
  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 add method

C. The head should be initialized to newNode inside the else block

D. The Node class should have a 'prev' pointer

Solution

1. Step 1: Analyze the while loop condition

   The loop currently runs while temp != null, which moves temp to null and loses reference to last node.

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.

3. Final Answer:

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

4. Quick Check:

   Loop stops at last node using temp.next != null [OK]

Hint: Loop to last node uses 'temp.next != null' [OK]

Common Mistakes:

• Looping until temp is null loses last node reference
• Creating newNode outside add method is unnecessary
• Adding 'prev' pointer in singly linked list node

5. You want to create a singly linked list from scratch that supports adding nodes at the head and printing the list. Which code snippet correctly implements adding at the head?

hard

A. public void addAtHead(int data) { Node newNode = new Node(data); head = newNode; newNode.next = head; }

B. public void addAtHead(int data) { Node newNode = new Node(data); head.next = newNode; head = newNode; }

C. public void addAtHead(int data) { Node newNode = new Node(data); newNode.next = head; head = newNode; }

D. public void addAtHead(int data) { Node newNode = new Node(data); newNode = head; head = newNode; }

Solution

1. Step 1: Understand adding at head logic

   To add at head, new node's next should point to current head, then head updated to new node.

2. Step 2: Check each option's correctness

   public void addAtHead(int data) { Node newNode = new Node(data); newNode.next = head; head = newNode; } correctly sets newNode.next = head and then head = newNode. Others incorrectly assign or reverse pointers.

3. Final Answer:

   public void addAtHead(int data) { Node newNode = new Node(data); newNode.next = head; head = newNode; } -> Option C

4. Quick Check:

   Add at head = newNode.next = head; head = newNode [OK]

Hint: New head points to old head, then update head [OK]

Common Mistakes:

• Assigning head.next before updating head
• Reversing pointer assignments
• Overwriting newNode reference incorrectly