Linked List vs Array When to Choose Which in DSA Rust - Build Both Approaches

Linked List vs Array: When to Choose Which

📖 Scenario: Imagine you are managing a playlist of songs on a music app. You want to understand when to use an array or a linked list to store your songs. Arrays let you quickly find songs by their position, but adding or removing songs in the middle can be slow. Linked lists let you add or remove songs easily anywhere, but finding a song by position takes longer.

🎯 Goal: You will create a simple Rust program to store a list of song names first using an array (vector) and then using a linked list. You will see how to add songs and print the list in both cases. This will help you understand when to choose arrays or linked lists.

📋 What You'll Learn

Create a vector called playlist_array with 3 songs: "Song A", "Song B", "Song C"

Create a linked list called playlist_linked and add the same 3 songs in order

Add a new song "Song D" to the end of both the vector and the linked list

Print both playlists to see the order of songs

💡 Why This Matters

🌍 Real World

Music apps, video playlists, and other ordered collections often use arrays or linked lists depending on how users add, remove, or access items.

💼 Career

Understanding when to use arrays or linked lists helps in writing efficient code for software development, especially in systems programming and application development.

1

Create a vector playlist with 3 songs

Create a vector called playlist_array with these exact songs in order: "Song A", "Song B", "Song C"

DSA Rust

# Create a vector called playlist_array with 3 songs
# Your code here

Hint

Use vec!["Song A", "Song B", "Song C"] to create the vector.

2

Create a linked list playlist with the same 3 songs

Add code to create a linked list called playlist_linked and add the songs "Song A", "Song B", and "Song C" in order using push_back

DSA Rust

fn main() {
    let mut playlist_array = vec!["Song A", "Song B", "Song C"];
    // Create a linked list called playlist_linked and add 3 songs
    // Your code here
}

Hint

Use LinkedList::new() to create the list and push_back to add songs.

3

Add a new song "Song D" to both playlists

Add the song "Song D" to the end of playlist_array using push and to playlist_linked using push_back

DSA Rust

use std::collections::LinkedList;

fn main() {
    let mut playlist_array = vec!["Song A", "Song B", "Song C"];
    let mut playlist_linked = LinkedList::new();
    playlist_linked.push_back("Song A");
    playlist_linked.push_back("Song B");
    playlist_linked.push_back("Song C");
    // Add "Song D" to playlist_array and playlist_linked
    // Your code here
}

Hint

Use push for the vector and push_back for the linked list.

4

Print both playlists to see the songs

Print playlist_array and playlist_linked using println! to show all songs in order

DSA Rust

use std::collections::LinkedList;

fn main() {
    let mut playlist_array = vec!["Song A", "Song B", "Song C"];
    let mut playlist_linked = LinkedList::new();
    playlist_linked.push_back("Song A");
    playlist_linked.push_back("Song B");
    playlist_linked.push_back("Song C");
    playlist_array.push("Song D");
    playlist_linked.push_back("Song D");
    // Print playlist_array and playlist_linked
    // Your code here
}

Hint

Use println!("Playlist Array: {:?}", playlist_array); and similarly for the linked list.

Practice

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

easy

A. Queue

B. Array

C. Stack

D. Linked List

Solution

1. Step 1: Understand random access in arrays

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

2. Step 2: Compare with linked lists

   Linked lists require traversal from the head to reach an element, making random access slower.

3. Final Answer:

   Array -> Option B

4. Quick Check:

   Random access = Array [OK]

Hint: For fast index access, choose arrays [OK]

Common Mistakes:

• Thinking linked lists allow fast random access
• Confusing stacks or queues with arrays
• Assuming linked lists store elements contiguously

2. Which of the following is the correct Rust syntax to declare a mutable vector (dynamic array)?

easy

A. let mut v = vec[];

B. let v = Vec::new();

C. let mut v = Vec::new();

D. let v mut = Vec::new();

Solution

1. Step 1: Identify correct mutable vector declaration

   In Rust, 'let mut v = Vec::new();' declares a mutable vector named v.

2. Step 2: Check other options for syntax errors

   let v = Vec::new(); is immutable vector, C has wrong vec syntax, D has incorrect keyword order.

3. Final Answer:

   let mut v = Vec::new(); -> Option C

4. Quick Check:

   Mutable vector syntax = let mut v = Vec::new(); [OK]

Hint: Use 'let mut' before variable for mutability [OK]

Common Mistakes:

• Forgetting 'mut' for mutable vectors
• Using incorrect vec macro syntax
• Placing 'mut' after variable name

3. Consider this Rust code using a linked list from the standard library:

use std::collections::LinkedList;

let mut list = LinkedList::new();
list.push_back(10);
list.push_back(20);
list.push_front(5);

let first = list.pop_front();
println!("{:?}", first);
println!("{:?}", list);

What is the printed output?

medium

A. Some(5)
[10, 20]

B. Some(10)
[5, 20]

C. None
[5, 10, 20]

D. Some(5)
[5, 10, 20]

Solution

1. Step 1: Trace list operations

   Initially empty list. push_back(10) -> [10], push_back(20) -> [10, 20], push_front(5) -> [5, 10, 20].

2. Step 2: pop_front() removes and returns first element

   pop_front() returns Some(5), list becomes [10, 20].

3. Final Answer:

   Some(5) [10, 20] -> Option A

4. Quick Check:

   pop_front returns first element = Some(5) [OK]

Hint: pop_front removes first element from linked list [OK]

Common Mistakes:

• Confusing push_front and push_back order
• Expecting pop_front to return None
• Misreading list contents after operations

4. The following Rust code tries to remove an element from a linked list but causes a compile error:

use std::collections::LinkedList;

let mut list = LinkedList::new();
list.push_back(1);
list.push_back(2);
let removed = list.remove(0);

What is the main reason for the error?

medium

A. LinkedList in Rust does not have a remove method with index parameter

B. You cannot push_back elements into a LinkedList

C. The list must be immutable to remove elements

D. The remove method requires a mutable reference to the element

Solution

1. Step 1: Check LinkedList API in Rust

   Rust's std::collections::LinkedList does not provide a remove method that takes an index.

2. Step 2: Understand error cause

   Trying to call remove(0) causes compile error because such method does not exist.

3. Final Answer:

   LinkedList in Rust does not have a remove method with index parameter -> Option A

4. Quick Check:

   LinkedList remove by index = no method [OK]

Hint: Rust LinkedList lacks remove by index method [OK]

Common Mistakes:

• Assuming LinkedList supports remove by index
• Confusing mutability rules
• Thinking push_back is invalid

5. You need to store a large collection of elements where insertions and deletions happen frequently at both ends, but you also need to access elements by index occasionally. Which data structure is the best choice in Rust?

hard

A. LinkedList

B. HashMap

C. Vec (Array)

D. VecDeque

Solution

1. Step 1: Analyze requirements for insertions/deletions at both ends

   LinkedList and VecDeque both support fast insertions/removals at front and back.

2. Step 2: Consider occasional index access

   LinkedList has slow index access (O(n)), VecDeque supports efficient indexing (O(1) amortized).

3. Step 3: Compare Vec and HashMap

   Vec has slow front insertions, HashMap is for key-value, not ordered sequence.

4. Final Answer:

   VecDeque -> Option D

5. Quick Check:

   Fast ends + index access = VecDeque [OK]

Hint: Use VecDeque for fast ends and index access [OK]

Common Mistakes:

• Choosing LinkedList despite slow index access
• Choosing Vec despite slow front insertions
• Confusing HashMap with sequence structures