Creating a Singly Linked List from Scratch in DSA Typescript - Code Step by Step

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the Node class with a constructor that initializes the value and next properties.

DSA Typescript

class Node {
  value: number;
  next: Node | null;

  constructor(value: number) {
    this.value = [1];
    this.next = null;
  }
}

Drag options to blanks, or click blank then click option'

Avalue

Bnext

Cnull

Dundefined

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning 'this.next' instead of 'this.value'.

Assigning 'null' or 'undefined' to 'this.value'.

The constructor parameter 'value' should be assigned to the instance property 'this.value'.

2fill in blank

medium

Complete the code to initialize the head property of the LinkedList class to null.

DSA Typescript

class LinkedList {
  head: Node | null;

  constructor() {
    this.head = [1];
  }
}

Drag options to blanks, or click blank then click option'

Aundefined

Bnew Node(0)

C0

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'undefined' instead of 'null'.

Initializing head with a new Node instead of null.

The head of an empty linked list should be initialized to null.

3fill in blank

hard

Fix the error in the add method to correctly add a new node at the end of the list.

DSA Typescript

add(value: number): void {
  const newNode = new Node(value);
  if (this.head === null) {
    this.head = newNode;
  } else {
    let current = this.head;
    while (current.next !== [1]) {
      current = current.next;
    }
    current.next = newNode;
  }
}

Drag options to blanks, or click blank then click option'

Aundefined

B0

Cnull

DnewNode

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for 'undefined' instead of 'null'.

Comparing to '0' or 'newNode' incorrectly.

The loop should continue until current.next is null, meaning the end of the list.

4fill in blank

hard

Fill both blanks to complete the printList method that prints all node values separated by arrows.

DSA Typescript

printList(): void {
  let current = this.head;
  let result = '';
  while (current !== [1]) {
    result += current.value + [2];
    current = current.next;
  }
  console.log(result + 'null');
}

Drag options to blanks, or click blank then click option'

Anull

B' -> '

C' ->'

Dundefined

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'undefined' instead of 'null' in the loop condition.

Missing spaces around the arrow string.

The loop runs while current is not null, and values are joined with ' -> ' arrows.

5fill in blank

hard

Fill all three blanks to complete the remove method that deletes the first node with the given value.

DSA Typescript

remove(value: number): void {
  if (this.head === [1]) {
    return;
  }
  if (this.head.value === value) {
    this.head = this.head.next;
    return;
  }
  let current = this.head;
  while (current !== null && current.next !== [2]) {
    if (current.next.value === value) {
      current.next = current.next.[3];
      return;
    }
    current = current.next;
  }
}

Drag options to blanks, or click blank then click option'

Anull

Bundefined

Cnext

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'undefined' instead of 'null' for checks.

Assigning current.next to 'head' or 'undefined' incorrectly.

Check if head is null (return early), then check if head.value === value (remove head if match), then traverse while current.next !== null, and update current.next to current.next.next to skip the removed node.

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