Why Doubly Linked List Over Singly Linked List in DSA Rust - Challenge Your Understanding

Challenge - 5 Problems

🧠 Conceptual

intermediate

Advantages of Doubly Linked List

Which of the following is a key advantage of a doubly linked list compared to a singly linked list?

A It uses less memory per node than singly linked list.

B It allows traversal in both forward and backward directions.

C It requires fewer pointers per node.

D It can only be traversed in one direction.

💡 Hint

Think about how nodes are connected in both lists.

A doubly linked list has two pointers per node, allowing traversal forwards and backwards. A singly linked list has only one pointer per node, so it can only be traversed forward.

❓ Predict Output

intermediate

Output of Traversal in Doubly Linked List

What is the output of the following Rust code that traverses a doubly linked list backwards after inserting nodes 1, 2, 3?

DSA Rust

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

fn main() {
    // Simplified representation
    let list = vec![1, 2, 3];
    // Backward traversal output
    let output = list.iter().rev().map(|v| v.to_string()).collect::<Vec<_>>().join(" -> ");
    println!("{} -> null", output);
}

A 3 -> null

B 1 -> 2 -> 3 -> null

C 3 -> 2 -> 1 -> null

D null -> 1 -> 2 -> 3

💡 Hint

Backward traversal reverses the order of nodes.

The code reverses the vector representing the list and prints the values from last to first, resulting in "3 -> 2 -> 1 -> null".

❓ Predict Output

advanced

Effect of Removing Node in Doubly Linked List

Given a doubly linked list with nodes 1 <-> 2 <-> 3, what is the printed list after removing node 2?

DSA Rust

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

fn main() {
    let mut list = vec![1, 2, 3];
    list.remove(1); // remove node with value 2
    let output = list.iter().map(|v| v.to_string()).collect::<Vec<_>>().join(" -> ");
    println!("{} -> null", output);
}

A 1 -> 3 -> null

B 1 -> 2 -> 3 -> null

C 2 -> 3 -> null

D 3 -> 1 -> null

💡 Hint

Removing node 2 should link node 1 directly to node 3.

After removing node 2, nodes 1 and 3 become neighbors, so the list prints as "1 -> 3 -> null".

🔧 Debug

advanced

Identify the Bug in Doubly Linked List Node Insertion

What error will the following Rust code produce when inserting a new node at the front of a doubly linked list?

DSA Rust

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

fn insert_front(mut head: Option<Box<Node>>, val: i32) -> Option<Box<Node>> {
    let mut new_node = Box::new(Node { value: val, prev: None, next: head });
    if let Some(ref mut old_head) = new_node.next {
        old_head.prev = Some(new_node);
    }
    Some(new_node)
}

A error: cannot move out of `new_node` because it is borrowed

B no error, code runs correctly

C error: mismatched types between Option<Box<Node>> and Box<Node>

D error: use of moved value `head`

💡 Hint

Look at the use of `head` after assigning it to `new_node.next`.

The `head` parameter is moved into `new_node.next`, so attempting to destructure it in the `if let` results in a "use of moved value `head`" error.

🧠 Conceptual

expert

Memory Overhead of Doubly Linked List

Why does a doubly linked list generally use more memory than a singly linked list?

A Because each node stores two pointers instead of one, increasing memory per node.

B Because doubly linked lists store data twice in each node.

C Because doubly linked lists require extra memory for garbage collection.

D Because doubly linked lists use arrays internally.

💡 Hint

Consider what extra information each node holds.

Doubly linked list nodes have two pointers (prev and next), so each node uses more memory than singly linked list nodes which have only one pointer.

Practice

1. Why might a doubly linked list be preferred over a singly linked list in some cases?

easy

A. Because it cannot be used to implement stacks

B. Because it allows traversal in both forward and backward directions

C. Because it only stores one pointer per node

D. Because it uses less memory than singly linked list

Solution

1. Step 1: Understand traversal capabilities

   A singly linked list allows traversal only in one direction, from head to tail.

2. Step 2: Compare with doubly linked list

   A doubly linked list stores two pointers per node, allowing traversal both forward and backward.

3. Final Answer:

   Because it allows traversal in both forward and backward directions -> Option B

4. Quick Check:

   Traversal direction = Both forward and backward [OK]

Hint: Doubly linked lists have two pointers per node [OK]

Common Mistakes:

• Thinking doubly linked lists use less memory
• Confusing pointer count per node
• Assuming singly linked lists allow backward traversal

2. Which of the following is the correct way to declare a node struct for a doubly linked list in Rust?

easy

A. struct Node { data: i32, prev: Option, next: Option }

B. struct Node { data: i32, prev: Box<Node> }

C. struct Node { data: i32, next: Option }

D. struct Node { data: i32, next: Node }

Solution

1. Step 1: Identify doubly linked list node fields

   A doubly linked list node must have pointers to both previous and next nodes, usually wrapped in Option and Box for ownership.

2. Step 2: Check each option

   struct Node { data: i32, prev: Option, next: Option } correctly declares both prev and next as Option. Others miss prev or use incorrect types.

3. Final Answer:

   struct Node { data: i32, prev: Option, next: Option } -> Option A

4. Quick Check:

   Prev and next pointers = struct Node { data: i32, prev: Option, next: Option } [OK]

Hint: Doubly linked nodes have prev and next pointers [OK]

Common Mistakes:

• Omitting prev pointer
• Using Box<Node> without Option
• Using next as Node instead of pointer

3. Given this Rust code snippet for a doubly linked list insertion at the end, what will be the printed list after inserting 3 nodes with values 10, 20, 30?

let mut list = DoublyLinkedList::new();
list.push_back(10);
list.push_back(20);
list.push_back(30);
list.print_forward();

medium

A. 10 -> 30 -> 20 -> null

B. 30 -> 20 -> 10 -> null

C. null -> 10 -> 20 -> 30

D. 10 -> 20 -> 30 -> null

Solution

1. Step 1: Understand push_back behavior

   push_back adds nodes at the end, so nodes are appended in order 10, then 20, then 30.

2. Step 2: Trace print_forward output

   Printing forward shows nodes from head to tail: 10 -> 20 -> 30 -> null.

3. Final Answer:

   10 -> 20 -> 30 -> null -> Option D

4. Quick Check:

   Insertion order = Printed order [OK]

Hint: push_back appends at tail, print_forward shows head to tail [OK]

Common Mistakes:

• Confusing push_back with push_front
• Printing backward instead of forward
• Mixing node order in output

4. In this Rust doubly linked list code, what is the bug causing incorrect backward traversal?

fn print_backward(&self) {
    let mut current = self.tail.clone();
    while let Some(node) = current {
        print!("{} -> ", node.data);
        current = node.next.clone();
    }
    println!("null");
}

medium

A. Not cloning the tail pointer

B. Printing node.data instead of node.prev

C. Using node.next instead of node.prev for backward traversal

D. Missing a break statement in the loop

Solution

1. Step 1: Analyze traversal direction

   Backward traversal should follow prev pointers, but code uses node.next which goes forward.

2. Step 2: Identify correct pointer usage

   Replacing node.next with node.prev fixes backward traversal.

3. Final Answer:

   Using node.next instead of node.prev for backward traversal -> Option C

4. Quick Check:

   Backward traversal = Use prev pointer [OK]

Hint: Backward traversal uses prev pointer, not next [OK]

Common Mistakes:

• Confusing next and prev pointers
• Not cloning pointers properly
• Printing wrong data fields

5. You want to implement a feature that allows quick removal of a node from the middle of a list without traversing from the head. Why is a doubly linked list better than a singly linked list for this task?

hard

A. Because doubly linked list nodes have pointers to previous nodes, enabling direct access to neighbors

B. Because singly linked lists store extra data for quick removal

C. Because doubly linked lists use less memory, making removal faster

D. Because singly linked lists allow backward traversal

Solution

1. Step 1: Understand node removal requirements

   Removing a node requires updating pointers of its neighbors, including the previous node.

2. Step 2: Compare singly and doubly linked lists

   Singly linked lists lack backward pointers, so to remove a node, you must traverse from head to find the previous node. Doubly linked lists have prev pointers, allowing direct access to neighbors.

3. Final Answer:

   Because doubly linked list nodes have pointers to previous nodes, enabling direct access to neighbors -> Option A

4. Quick Check:

   Prev pointer enables quick removal [OK]

Hint: Prev pointer lets you remove nodes without full traversal [OK]

Common Mistakes:

• Assuming singly linked lists have backward pointers
• Thinking memory usage affects removal speed
• Confusing traversal direction capabilities