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

Time Complexity: Creating a Singly Linked List from Scratch

O(n)

Understanding Time Complexity

When we build a singly linked list from scratch, we want to know how the time it takes grows as we add more items.

We ask: How does adding nodes affect the total work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class Node {
  value: number;
  next: Node | null = null;
  constructor(value: number) {
    this.value = value;
  }
}

class SinglyLinkedList {
  head: Node | null = null;

  append(value: number) {
    const newNode = new Node(value);
    if (!this.head) {
      this.head = newNode;
      return;
    }
    let current = this.head;
    while (current.next) {
      current = current.next;
    }
    current.next = newNode;
  }
}

This code creates a singly linked list and adds new nodes to the end.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves through nodes to find the list end.
• How many times: It runs once for each node already in the list before adding the new one.

How Execution Grows With Input

As the list grows, adding a new node takes longer because we must walk through all existing nodes.

Input Size (n)	Approx. Operations
10	About 10 steps to add the 10th node
100	About 100 steps to add the 100th node
1000	About 1000 steps to add the 1000th node

Pattern observation: The work grows linearly with the number of nodes already in the list.

Final Time Complexity

Time Complexity: O(n)

This means adding a new node takes time proportional to the current list size.

Common Mistake

[X] Wrong: "Adding a node always takes the same time, no matter the list size."

[OK] Correct: Because we must walk through all nodes to find the end, adding takes longer as the list grows.

Interview Connect

Understanding this helps you explain how linked lists work and why some operations are slower than others, a key skill in coding interviews.

Self-Check

"What if we kept a pointer to the last node? How would the time complexity of adding a node 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 one

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-like hierarchy

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 in the sequence.

2. Step 2: Compare with other data structures

   Unlike key-value stores or trees, singly linked lists maintain a simple linear chain of elements.

3. Final Answer:

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

4. Quick Check:

   Singly Linked List = sequence with next pointers [OK]

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

Common Mistakes:

• Confusing linked lists with arrays
• Thinking linked lists store key-value pairs
• Assuming linked lists are hierarchical like trees

2. Which of the following is the correct TypeScript syntax to define a node class for a singly linked list?

easy

A. class Node { value: number; next: Node | null; constructor(value: number) { this.value = value; this.next = null; } }

B. class Node { value: number; next: Node; constructor(value: number) { this.value = value; this.next = null; } }

C. class Node { value: number; next: null; constructor(value: number) { this.value = value; this.next = null; } }

D. class Node { value: number; next: any; constructor(value: number) { this.value = value; this.next = undefined; } }

Solution

1. Step 1: Check the type of the next pointer

   The next pointer should be either another Node or null if it is the last node.

2. Step 2: Verify constructor initialization

   The constructor sets the value and initializes next to null, which is correct for a new node.

3. Final Answer:

   class Node { value: number; next: Node | null; constructor(value: number) { this.value = value; this.next = null; } } -> Option A

4. Quick Check:

   next can be Node or null = class Node { value: number; next: Node | null; constructor(value: number) { this.value = value; this.next = null; } } [OK]

Hint: Use union type for next: Node | null [OK]

Common Mistakes:

• Forgetting to allow next to be null
• Setting next to undefined instead of null
• Using any type for next which loses type safety

3. What will be the output after running this TypeScript code that creates a singly linked list with values 10, 20, 30 and prints it?

class Node {
  value: number;
  next: Node | null;
  constructor(value: number) {
    this.value = value;
    this.next = null;
  }
}

class LinkedList {
  head: Node | null = null;

  append(value: number) {
    const newNode = new Node(value);
    if (!this.head) {
      this.head = newNode;
      return;
    }
    let current = this.head;
    while (current.next) {
      current = current.next;
    }
    current.next = newNode;
  }

  print() {
    let current = this.head;
    let result = '';
    while (current) {
      result += current.value + ' -> ';
      current = current.next;
    }
    result += 'null';
    console.log(result);
  }
}

const list = new LinkedList();
list.append(10);
list.append(20);
list.append(30);
list.print();

medium

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

B. null

C. 10 -> 20 -> null

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

Solution

1. Step 1: Append nodes in order

   Nodes with values 10, 20, and 30 are appended in that order, so the list is 10 -> 20 -> 30.

2. Step 2: Print the list

   The print method traverses from head to end, concatenating values with ' -> ' and ending with 'null'.

3. Final Answer:

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

4. Quick Check:

   Appending preserves order = 10 -> 20 -> 30 -> null [OK]

Hint: Append adds at end; print shows sequence [OK]

Common Mistakes:

• Assuming append adds at head instead of tail
• Forgetting to add 'null' at the end
• Printing values in reverse order

4. Identify the error in this TypeScript code snippet that tries to append a node to a singly linked list:

class Node {
  value: number;
  next: Node | null;
  constructor(value: number) {
    this.value = value;
    this.next = null;
  }
}

class LinkedList {
  head: Node | null = null;

  append(value: number) {
    const newNode = new Node(value);
    if (this.head === null) {
      this.head = newNode;
    } else {
      let current = this.head;
      while (current.next !== null) {
        current = current.next;
      }
      current.next = newNode;
    }
  }
}

const list = new LinkedList();
list.append(5);
list.append(10);
console.log(list.head.next.value);

medium

A. Error: head is undefined

B. Error: Cannot read property 'value' of null

C. No error; output will be 10

D. Error: next property is missing in Node

Solution

1. Step 1: Check append logic

   The append method correctly adds the first node as head and subsequent nodes at the end.

2. Step 2: Verify output of console.log

   After appending 5 and 10, head.next points to the node with value 10, so accessing value is valid.

3. Final Answer:

   No error; output will be 10 -> Option C

4. Quick Check:

   Proper append and access = 10 [OK]

Hint: Check if head and next are properly assigned before access [OK]

Common Mistakes:

• Forgetting to return after setting head in append
• Accessing next.value when next is null
• Not initializing next to null in Node

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 links between nodes?

hard

A. Create a new list and append nodes from the original list in reverse order

B. Iterate through the list, for each node set its next to the previous node, updating previous and current pointers

C. Swap the values of the first and last nodes repeatedly until the middle is reached

D. Remove all nodes and re-insert them at the head one by one

Solution

1. Step 1: Understand in-place reversal

   Reversing links means changing each node's next pointer to point to the previous node, not creating new nodes.

2. Step 2: Identify correct method

   Iterating with previous and current pointers and updating next links reverses the list efficiently in place.

3. Final Answer:

   Iterate through the list, for each node set its next to the previous node, updating previous and current pointers -> Option B

4. Quick Check:

   In-place reversal = update next pointers backward [OK]

Hint: Reverse by redirecting next pointers, not swapping values [OK]

Common Mistakes:

• Swapping node values instead of links
• Creating a new list instead of reversing in place
• Removing and re-adding nodes unnecessarily