Why Linked List Exists and What Problem It Solves in DSA Typescript - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node in a linked list.

DSA Typescript

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

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

A0

Bundefined

Cthis

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Setting next to undefined or 0 instead of null

Setting next to this which causes circular reference

When creating a new node, its next pointer should start as null because it does not point to any other node yet.

2fill in blank

medium

Complete the code to add a new node at the beginning of the linked list.

DSA Typescript

class LinkedList {
  head: Node | null = null;
  addFirst(value: number) {
    const newNode = new Node(value);
    newNode.next = [1];
    this.head = newNode;
  }
}

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

AnewNode

Bthis.head.next

Cthis.head

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Setting newNode.next to null loses the rest of the list

Setting newNode.next to newNode causes a cycle

To add a node at the start, the new node's next should point to the current head, then update head to new node.

3fill in blank

hard

Fix the error in the code that tries to traverse the linked list to print all values.

DSA Typescript

printList(head: Node | null) {
  let current = head;
  while (current [1] null) {
    console.log(current.value);
    current = current.next;
  }
}

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

A!==

B==

C=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using assignment (=) instead of comparison

Using == which can cause unexpected type coercion

The loop should continue while current is not null, so use !== null.

4fill in blank

hard

Fill both blanks to create a linked list from an array of numbers.

DSA Typescript

function createLinkedList(arr: number[]): Node | null {
  let head: Node | null = null;
  for (let i = arr.length - 1; i >= 0; i--) {
    const newNode = new Node(arr[i]);
    newNode.next = [1];
    head = [2];
  }
  return head;
}

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

Ahead

BnewNode

Carr[i]

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Setting newNode.next to arr[i] which is a number, not a node

Not updating head to newNode

We set newNode.next to the current head, then update head to newNode to build the list from the end.

5fill in blank

hard

Fill all three blanks to remove the first node from the linked list.

DSA Typescript

class LinkedList {
  head: Node | null = null;
  removeFirst() {
    if (this.head !== null) {
      const removed = this.head;
      this.head = [1];
      removed.next = [2];
      return [3];
    }
    return null;
  }
}

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

Athis.head.next

Bnull

Cremoved

Attempts:

3 left

💡 Hint

Common Mistakes

Not updating head to head.next

Not disconnecting removed node's next

Returning head instead of removed node

To remove the first node, update head to head.next, disconnect removed node's next by setting it to null, and return the removed node.

Practice

1. Why do we use a linked list instead of an array in some cases?

easy

A. Because linked lists allow easy insertion and deletion without shifting elements

B. Because linked lists use less memory than arrays always

C. Because linked lists store elements in contiguous memory locations

D. Because linked lists are faster for accessing elements by index

Solution

1. Step 1: Understand array limitations

   Arrays store elements in continuous memory, so inserting or deleting in the middle requires shifting many elements, which is slow.

2. Step 2: Understand linked list advantage

   Linked lists store elements as nodes with pointers, so insertion or deletion only changes pointers without shifting elements.

3. Final Answer:

   Because linked lists allow easy insertion and deletion without shifting elements -> Option A

4. Quick Check:

   Linked list insertion/deletion = easy [OK]

Hint: Linked lists avoid shifting elements on insert/delete [OK]

Common Mistakes:

• Thinking linked lists always use less memory
• Believing linked lists store elements contiguously
• Assuming linked lists have fast random access

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

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 next pointer type

   In a singly linked list, next can be another Node or null if it's the last node.

2. Step 2: Verify constructor initialization

   The constructor sets value and initializes next to null correctly.

3. Final Answer:

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

4. Quick Check:

   Node next type = Node or null [OK]

Hint: Next pointer must allow null for last node [OK]

Common Mistakes:

• Not allowing next to be null
• Setting next to undefined instead of null
• Using any type for next pointer

3. What will be the output of the following TypeScript code that uses a linked list?

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

const head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);

let current = head;
const values: number[] = [];
while (current !== null) {
  values.push(current.value);
  current = current.next;
}
console.log(values);

medium

A. []

B. [3, 2, 1]

C. [1, 2]

D. [1, 2, 3]

Solution

1. Step 1: Build linked list nodes

   Nodes created with values 1, 2, 3 linked in order: 1 -> 2 -> 3 -> null.

2. Step 2: Traverse linked list and collect values

   Loop visits nodes in order: 1, then 2, then 3, pushing values into array.

3. Final Answer:

   [1, 2, 3] -> Option D

4. Quick Check:

   Traversal order = [1, 2, 3] [OK]

Hint: Traverse from head to null to get values in order [OK]

Common Mistakes:

• Confusing order and reversing output
• Stopping traversal too early
• Not initializing next pointers correctly

4. The following TypeScript code tries to insert a new node with value 4 at the end of the linked list but has a bug. What is the bug?

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

const head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);

function insertAtEnd(head: Node, value: number) {
  let current = head;
  while (current.next !== null) {
    current = current.next;
  }
  current = new Node(value);
}

insertAtEnd(head, 4);

let current = head;
const values: number[] = [];
while (current !== null) {
  values.push(current.value);
  current = current.next;
}
console.log(values);

medium

A. The function should return the new head node

B. The new node is not linked to the last node; assignment should be current.next = new Node(value)

C. The while loop condition should be current !== null

D. The new node value should be assigned before the loop

Solution

1. Step 1: Analyze insertion logic

   The code moves current to last node, but then assigns current = new Node(value), which only changes local variable, not the list.

2. Step 2: Correct pointer assignment

   To link new node, must assign current.next = new Node(value) so last node points to new node.

3. Final Answer:

   The new node is not linked to the last node; assignment should be current.next = new Node(value) -> Option B

4. Quick Check:

   Link new node with current.next = new Node(value) [OK]

Hint: Assign new node to current.next, not current [OK]

Common Mistakes:

• Assigning new node to current variable only
• Wrong loop condition causing null errors
• Expecting function to return new head unnecessarily

5. You want to store a list of tasks where you often add or remove tasks from the middle. Which data structure is best and why? A linked list or an array? Choose the best option.

hard

A. Array, because it uses less memory than linked lists

B. Array, because it allows fast random access to any task by index

C. Linked list, because it allows efficient insertion and deletion in the middle without shifting elements

D. Linked list, because it stores elements in contiguous memory for cache efficiency

Solution

1. Step 1: Understand operation needs

   Adding/removing tasks often in the middle means many insertions and deletions not at the end.

2. Step 2: Compare array vs linked list for middle operations

   Arrays require shifting elements on insert/delete in middle, which is slow. Linked lists only change pointers, so faster for these operations.

3. Final Answer:

   Linked list, because it allows efficient insertion and deletion in the middle without shifting elements -> Option C

4. Quick Check:

   Middle insert/delete = linked list best [OK]

Hint: For frequent middle changes, linked list is faster [OK]

Common Mistakes:

• Choosing array for fast random access ignoring insert/delete cost
• Thinking linked lists use less memory always
• Believing linked lists store elements contiguously