Memory Layout Comparison Array vs Linked List in DSA Rust - Trade-offs & Analysis

Mental Model

Arrays store items next to each other in memory, while linked lists store items scattered with pointers connecting them.

Analogy: Imagine a row of houses on a street (array) versus houses spread across a neighborhood connected by roads (linked list).

Array: [1][2][3][4][5]
Linked List: 1 -> 2 -> 3 -> 4 -> 5 -> null

Dry Run Walkthrough

Input: Compare memory layout of array [1, 2, 3] and linked list 1 -> 2 -> 3 -> null

Goal: Understand how elements are stored and connected in memory for both structures

Step 1: Visualize array elements stored contiguously

[1][2][3]

Why: Array elements are placed side by side in memory for fast access

Step 2: Visualize linked list nodes scattered with pointers

1 -> 2 -> 3 -> null

Why: Linked list nodes can be anywhere in memory, connected by pointers

Step 3: Note array memory addresses increase by fixed size

Addr1, Addr2, Addr3 (contiguous)

Why: Array elements are stored in consecutive memory blocks

Step 4: Note linked list nodes have pointers to next node's address

Node1(addrX) -> Node2(addrY) -> Node3(addrZ)

Why: Pointers link nodes even if scattered in memory

Result:

Array: [1][2][3] (contiguous memory)
Linked List: 1 -> 2 -> 3 -> null (nodes scattered, linked by pointers)

Annotated Code

DSA Rust

use std::ptr;

struct Node {
    value: i32,
    next: Option<Box<Node>>,
}

impl Node {
    fn new(value: i32) -> Self {
        Node { value, next: None }
    }
}

fn print_array(arr: &[i32]) {
    print!("Array: ");
    for val in arr {
        print!("[{}]", val);
    }
    println!();
}

fn print_linked_list(head: &Option<Box<Node>>) {
    print!("Linked List: ");
    let mut current = head;
    while let Some(node) = current {
        print!("{} -> ", node.value);
        current = &node.next;
    }
    println!("null");
}

fn main() {
    // Array example
    let arr = [1, 2, 3];
    print_array(&arr);

    // Linked list example
    let mut head = Box::new(Node::new(1));
    head.next = Some(Box::new(Node::new(2)));
    head.next.as_mut().unwrap().next = Some(Box::new(Node::new(3)));

    let head_option = Some(head);
    print_linked_list(&head_option);
}

for val in arr { print!("[{}]", val); }

Print array elements in order showing contiguous storage

while let Some(node) = current { print!("{} -> ", node.value); current = &node.next; }

Traverse linked list nodes following pointers to print values

Output

Array: [1][2][3]

Linked List: 1 -> 2 -> 3 -> null

Complexity Analysis

Time: O(n) because we visit each element once to print

Space: O(n) for storing n elements in both array and linked list

vs Alternative: Array uses contiguous memory for fast access but fixed size; linked list uses scattered memory with pointers allowing dynamic size but slower access

Edge Cases

Empty array and empty linked list

Both print nothing or indicate empty structure

DSA Rust

while let Some(node) = current { ... } handles empty linked list gracefully

When to Use This Pattern

When a problem involves understanding memory access speed or dynamic size, compare arrays and linked lists to choose the right structure.

Common Mistakes

Mistake: Assuming linked list elements are stored next to each other in memory
Fix: Remember linked list nodes can be anywhere; they are connected by pointers, not by memory location

Summary

Shows how arrays store elements contiguously and linked lists store nodes scattered with pointers

Use arrays for fast indexed access and linked lists for flexible size with frequent insertions/deletions

The key insight is arrays use continuous memory blocks, linked lists use pointers to connect scattered nodes

Practice

1. Which statement best describes the memory layout difference between an array and a linked list?

easy

A. A linked list stores elements contiguously, but an array uses pointers between elements.

B. Both array and linked list store elements contiguously in memory.

C. Arrays use pointers to link elements, linked lists store elements contiguously.

D. An array stores elements contiguously in memory, while a linked list stores elements scattered with pointers.

Solution

1. Step 1: Understand array memory layout

   An array stores all its elements in a continuous block of memory, so elements are next to each other.

2. Step 2: Understand linked list memory layout

   A linked list stores each element separately in memory, and each element has a pointer to the next, so elements are scattered.

3. Final Answer:

   An array stores elements contiguously in memory, while a linked list stores elements scattered with pointers. -> Option D

4. Quick Check:

   Array = contiguous, Linked list = scattered with pointers [OK]

Hint: Arrays are continuous blocks; linked lists use pointers [OK]

Common Mistakes:

• Thinking linked lists store elements contiguously
• Confusing pointers usage between arrays and linked lists
• Assuming arrays use pointers between elements

2. Which Rust code snippet correctly declares a fixed-size array of 5 integers with explicit type annotation?

easy

A. let arr = [1, 2, 3, 4, 5];

B. let arr = vec![1, 2, 3, 4, 5];

C. let arr: [i32; 5] = [1, 2, 3, 4, 5];

D. let arr = (1, 2, 3, 4, 5);

Solution

1. Step 1: Identify fixed-size array syntax in Rust

   Rust fixed-size arrays require type and size: [type; size].

2. Step 2: Check each option

   let arr: [i32; 5] = [1, 2, 3, 4, 5]; uses let arr: [i32; 5] = [1, 2, 3, 4, 5]; which is correct fixed-size array syntax.

3. Final Answer:

   let arr: [i32; 5] = [1, 2, 3, 4, 5]; -> Option C

4. Quick Check:

   Fixed-size array syntax = [type; size] [OK]

Hint: Fixed arrays need type and size: [type; size] [OK]

Common Mistakes:

• Using vec! macro for fixed-size array
• Missing type and size in array declaration
• Using tuple syntax instead of array

3. Consider this Rust code creating a linked list node struct and two nodes linked together:

struct Node {
    value: i32,
    next: Option,
}

fn main() {
    let node2 = Node { value: 20, next: None };
    let node1 = Node { value: 10, next: Some(Box::new(node2)) };
    println!("{} -> {} -> None", node1.value, node1.next.as_ref().unwrap().value);
}

What is the printed output?

medium

A. 10 -> 20 -> None

B. 20 -> 10 -> None

C. 10 -> None -> None

D. Compilation error due to ownership

Solution

1. Step 1: Understand node creation and linking

   node2 has value 20 and no next node. node1 has value 10 and points to node2 via Box.

2. Step 2: Trace the print statement

   Prints node1.value (10), then node1.next's value (20), then "None" for end.

3. Final Answer:

   10 -> 20 -> None -> Option A

4. Quick Check:

   Linked nodes print values in order: 10 then 20 [OK]

Hint: Follow pointers to print linked list values [OK]

Common Mistakes:

• Confusing order of nodes
• Expecting None printed as value
• Thinking ownership causes compile error here

4. This Rust code tries to create a linked list but fails to compile:

struct Node {
    value: i32,
    next: Option,
}

fn main() {
    let node1 = Node { value: 10, next: None };
    let node2 = Node { value: 20, next: Some(Box::new(node1)) };
    println!("{} -> {} -> None", node1.value, node2.value);
}

What is the main reason for the compilation error?

medium

A. Use of moved value: node1 is moved into node2 and cannot be used again

B. Missing lifetime annotations on struct Node

C. Box cannot be used inside Option

D. Printing node2.value is invalid syntax

Solution

1. Step 1: Analyze ownership of node1

   node1 is moved into Box inside node2's next field, so node1 cannot be used after move.

2. Step 2: Identify error cause

   Because node1 is moved, any further use or move causes compile error "use of moved value".

3. Final Answer:

   Use of moved value: node1 is moved into node2 and cannot be used again -> Option A

4. Quick Check:

   Ownership moved node1 causes compile error [OK]

Hint: Moving ownership invalidates original variable [OK]

Common Mistakes:

• Thinking Box inside Option is invalid
• Assuming missing lifetimes cause error here
• Believing print syntax is wrong

5. You want to store 1 million integers and frequently access elements by index. Considering memory layout and performance, which data structure is better in Rust and why?

hard

A. Array, because linked lists cannot store integers efficiently.

B. Array, because elements are stored contiguously allowing fast index access and better cache usage.

C. Linked list, because it uses less memory than arrays for large data.

D. Linked list, because it uses pointers and is faster for random access.

Solution

1. Step 1: Consider access pattern and memory layout

   Arrays store elements contiguously, so accessing by index is O(1) and cache-friendly.

2. Step 2: Compare linked list access and memory use

   Linked lists store elements scattered with pointers, making index access O(n) and less cache efficient.

3. Final Answer:

   Array, because elements are stored contiguously allowing fast index access and better cache usage. -> Option B

4. Quick Check:

   Contiguous arrays = fast index access [OK]

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

Common Mistakes:

• Assuming linked lists are faster for index access
• Thinking linked lists use less memory for large data
• Believing arrays cannot store large data efficiently