Why Linked List Exists and What Problem It Solves in DSA Typescript - See It Work

Why Linked List Exists and What Problem It Solves

📖 Scenario: Imagine you have a collection of books that you want to organize on a shelf. Sometimes you want to add a new book in the middle without moving all the other books. Using a linked list helps you do this easily.

🎯 Goal: You will create a simple linked list of numbers to understand why linked lists exist and how they solve the problem of adding or removing items efficiently.

📋 What You'll Learn

Create a linked list node class with a value and a pointer to the next node

Create a linked list with three nodes containing values 10, 20, and 30

Add a new node with value 15 between nodes with values 10 and 20

Print the linked list values in order

💡 Why This Matters

🌍 Real World

Linked lists are used in music players to keep track of songs in a playlist where you can add or remove songs easily.

💼 Career

Understanding linked lists helps in software development roles where efficient data management and memory use are important.

1

Create Linked List Nodes

Create a class called ListNode with a constructor that takes a value of type number and a next pointer initialized to null.

DSA Typescript

// Your code here

Hint

Think of each node as a box holding a number and a link to the next box.

2

Create Initial Linked List

Create three ListNode objects called node1, node2, and node3 with values 10, 20, and 30 respectively. Link node1.next to node2 and node2.next to node3.

DSA Typescript

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

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

// Your code here

Hint

Link the nodes by setting the next property to the next node.

3

Insert a Node in the Middle

Create a new ListNode called node15 with value 15. Insert it between node1 and node2 by updating pointers: set node15.next to node2 and node1.next to node15.

DSA Typescript

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

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

const node1 = new ListNode(10);
const node2 = new ListNode(20);
const node3 = new ListNode(30);

node1.next = node2;
node2.next = node3;

// Your code here

Hint

To insert, point the new node to the next node, then point the previous node to the new node.

4

Print the Linked List

Create a variable current and set it to node1. Use a while loop to print each node's value followed by ' -> ' until current is null. After the loop, print 'null'.

DSA Typescript

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

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

const node1 = new ListNode(10);
const node2 = new ListNode(20);
const node3 = new ListNode(30);

node1.next = node2;
node2.next = node3;

const node15 = new ListNode(15);
node15.next = node2;
node1.next = node15;

// Your code here

Hint

Use a loop to follow the next pointers and build a string to print.

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