Creating a Singly Linked List from Scratch in DSA Typescript - Revision Summary

Recall & Review

beginner

What is a singly linked list?

A singly linked list is a chain of nodes where each node holds data and a reference to the next node. It starts at a head node and ends when a node points to null.

beginner

What does each node in a singly linked list contain?

Each node contains two parts: the data (value) and a pointer (reference) to the next node in the list.

intermediate

How do you add a new node at the end of a singly linked list?

To add a node at the end, create the new node, then find the last node (which points to null), and update its next pointer to the new node.

beginner

Why is a singly linked list different from an array?

A singly linked list stores elements in nodes connected by pointers, allowing easy insertion and deletion without shifting elements, unlike arrays which store elements in contiguous memory.

intermediate

What is the time complexity to access the nth element in a singly linked list?

Accessing the nth element requires traversing from the head node through n nodes, so the time complexity is O(n).

What does the 'next' pointer in a singly linked list node point to?

A The next node in the list

B The previous node in the list

C The head node

D The last node in the list

In a singly linked list, each node's 'next' pointer points to the next node in the sequence.

What value does the 'next' pointer of the last node in a singly linked list hold?

A It points to a random node

B It points to itself

C It points to the head node

D It is null (or None)

The last node's 'next' pointer is null, indicating the end of the list.

Which operation is easiest to perform in a singly linked list compared to an array?

A Inserting a node at the beginning

B Accessing the middle element

C Accessing an element by index

D Sorting the elements

Inserting at the beginning is easy because you just change the head pointer to the new node.

What is the first step to create a singly linked list from scratch?

A Create an array to hold nodes

B Sort the data

C Define a node structure with data and next pointer

D Initialize the tail pointer

You start by defining the node structure that holds data and a pointer to the next node.

If you want to traverse a singly linked list, where do you start?

A From the middle node

B From the head node

C From the tail node

D From any random node

Traversal always starts from the head node and moves through each next pointer.

Explain how to create a singly linked list from scratch and add a node to it.

Think about how nodes connect and how the head pointer changes.

Describe the difference between a singly linked list and an array in terms of memory and operations.

Consider how data is stored and accessed in both.

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