Linked List vs Array When to Choose Which in DSA Typescript - Why the Distinction Matters

The Big Idea

Discover why your favorite apps choose different ways to store lists and how it makes your experience smooth or slow!

The Scenario

Imagine you have a long list of your favorite songs written on paper. You want to add new songs, remove some, or find a song quickly. If you write them all in one long line (like an array), adding or removing songs in the middle means rewriting many songs. If you write each song on a separate card and link them with strings (like a linked list), you can add or remove cards easily without rewriting the whole list.

The Problem

Using just a simple list (array) means every time you add or remove a song in the middle, you must shift many songs up or down. This takes time and can cause mistakes if you lose track. Also, if your list grows beyond the paper size, you need a bigger paper and rewrite everything. This is slow and frustrating.

The Solution

Linked lists solve this by letting you connect each song card to the next with a string. You can add or remove cards by just changing the strings, without moving all cards. Arrays let you find songs quickly by position, but linked lists let you change the list easily without big rewrites.

Before vs After

✗ Before

let songs = ['Song1', 'Song2', 'Song3'];
songs.splice(1, 0, 'NewSong'); // shifts elements after index 1

✓ After

class Node {
  value: string;
  next: Node | null;
  constructor(value: string) {
    this.value = value;
    this.next = null;
  }
}
// Insert new node by changing links, no shifting needed

What It Enables

You can choose the best way to store and change your list depending on whether you need fast access or easy updates.

Real Life Example

Music apps use arrays to quickly play songs by number, but use linked lists to manage playlists where songs are often added or removed.

Key Takeaways

Arrays are great for quick access by position but slow for inserting/removing in the middle.

Linked lists are great for easy insertion/removal but slower to find items by position.

Choosing between them depends on whether you prioritize speed of access or ease of updates.

Practice

1. Which data structure is better when you need fast random access to elements by index?

easy

A. Stack

B. Linked List

C. Array

D. Queue

Solution

1. Step 1: Understand random access in arrays

   Arrays store elements in contiguous memory, so accessing by index is fast and direct.

2. Step 2: Understand random access in linked lists

   Linked lists require traversal from the head to reach an index, which is slower.

3. Final Answer:

   Array -> Option C

4. Quick Check:

   Fast random access = Array [OK]

Hint: Arrays give fast index access; linked lists do not [OK]

Common Mistakes:

• Thinking linked lists allow fast index access
• Confusing stack or queue with arrays
• Assuming all data structures have same access speed

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

easy

A. const node = new LinkedListNode(null);

B. const node = new LinkedListNode();

C. const node = LinkedListNode();

D. const node = LinkedListNode(null);

Solution

1. Step 1: Understand TypeScript class instantiation

   To create an instance, use the 'new' keyword with constructor parameters.

2. Step 2: Check constructor parameters

   LinkedListNode usually requires a value; null can represent empty data.

3. Final Answer:

   const node = new LinkedListNode(null); -> Option A

4. Quick Check:

   Use 'new' with constructor args for instances [OK]

Hint: Use 'new' keyword with constructor parameters to create nodes [OK]

Common Mistakes:

• Omitting 'new' keyword
• Calling class like a function without 'new'
• Passing no arguments when constructor expects one

3. Given this TypeScript code, what is the printed output?

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

medium

A. [1, 2, 3]

B. [3, 2, 1]

C. [1, 3, 2]

D. []

Solution

1. Step 1: Build the linked list nodes

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

2. Step 2: Traverse and collect values

   Loop pushes values 1, then 2, then 3 into result array.

3. Final Answer:

   [1, 2, 3] -> Option A

4. Quick Check:

   Traverse linked list in order = [1, 2, 3] [OK]

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

Common Mistakes:

• Confusing order of nodes
• Not updating current pointer correctly
• Expecting reversed output

4. This TypeScript code tries to insert a new node at the start of a linked list but has a bug. What is the error?

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

let head = new Node(1);
head.next = new Node(2);

function insertAtStart(newValue: number) {
  const newNode = new Node(newValue);
  newNode.next = head.next;
  head = newNode;
}

insertAtStart(0);

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

medium

A. head should not be reassigned inside the function

B. newNode.next should point to head, not head.next

C. Node constructor missing initialization of next

D. insertAtStart should return the new head

Solution

1. Step 1: Analyze insertion logic

   New node's next is set to head.next, skipping the current head node.

2. Step 2: Correct pointer assignment

   To insert at start, newNode.next must point to current head, preserving the list.

3. Final Answer:

   newNode.next should point to head, not head.next -> Option B

4. Quick Check:

   Insert at start = newNode.next = head [OK]

Hint: New head's next must link to old head, not old head's next [OK]

Common Mistakes:

• Setting newNode.next to head.next causes node loss
• Not updating head reference properly
• Confusing next pointers in linked list

5. You need to frequently insert and delete elements in the middle of a large collection. Which data structure is better and why? A) Array, because it allows fast random access. B) Linked List, because it supports binary search efficiently. C) Array, because it uses less memory. D) Linked List, because it allows fast insertion and deletion without shifting elements.

hard

A. Array, because it allows fast random access.

B. Linked List, because it supports binary search efficiently.

C. Array, because it uses less memory.

D. Linked List, because it allows fast insertion and deletion without shifting elements.

Solution

1. Step 1: Consider insertion and deletion in arrays

   Arrays require shifting elements when inserting or deleting in the middle, which is slow.

2. Step 2: Consider linked list advantages

   Linked lists can insert or delete nodes by changing pointers without shifting elements, making these operations faster.

3. Final Answer:

   Linked List, because it allows fast insertion and deletion without shifting elements. -> Option D

4. Quick Check:

   Frequent middle insert/delete = Linked List [OK]

Hint: Linked lists excel at middle insert/delete without shifting [OK]

Common Mistakes:

• Choosing array for frequent middle insert/delete
• Thinking linked lists support binary search efficiently
• Ignoring cost of shifting elements in arrays