Get Length of Linked List in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the length variable to zero before counting nodes.

DSA Rust

let mut length = [1];
let mut current = head.as_ref();
while let Some(node) = current {
    length += 1;
    current = node.next.as_ref();
}
println!("Length: {}", length);

Drag options to blanks, or click blank then click option'

A0

B1

CNone

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Starting length at 1 instead of 0

Using None or head as initial length

We start counting from zero because no nodes have been counted yet.

2fill in blank

medium

Complete the code to move to the next node in the linked list inside the loop.

DSA Rust

while let Some(node) = current {
    length += 1;
    current = node.[1].as_ref();
}

Drag options to blanks, or click blank then click option'

Aprev

Bparent

Cchild

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next'

Using unrelated field names like 'child' or 'parent'

We move to the next node by accessing the 'next' field of the current node.

3fill in blank

hard

Fix the error in the loop condition to correctly iterate over the linked list nodes.

DSA Rust

while let Some([1]) = current {
    length += 1;
    current = node.next.as_ref();
}

Drag options to blanks, or click blank then click option'

Anode

Bcurrent

Chead

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'current' which is the Option itself, not the inner value

Using unrelated variable names

The pattern variable should be 'node' to destructure the Option and access node fields.

4fill in blank

hard

Fill both blanks to define a function that returns the length of a linked list.

DSA Rust

fn get_length(head: Option<Box<Node>>) -> usize {
    let mut length = 0;
    let mut current = [1];
    while let Some(node) = current {
        length += 1;
        current = node.[2].as_ref();
    }
    length
}

Drag options to blanks, or click blank then click option'

Ahead.as_ref()

Bhead

Cnext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using head directly instead of head.as_ref()

Using 'prev' instead of 'next'

We start with a reference to the head node and move through 'next' pointers to count nodes.

5fill in blank

hard

Fill all three blanks to create a linked list node struct and a function to get its length.

DSA Rust

#[derive(Debug)]
struct Node {
    value: i32,
    [1]: Option<Box<Node>>,
}

fn get_length(head: Option<Box<Node>>) -> usize {
    let mut length = 0;
    let mut current = head.as_ref();
    while let Some(node) = current {
        length += 1;
        current = node.[2].as_ref();
    }
    length
}

fn main() {
    let node3 = Node { value: 3, [3]: None };
    println!("Length: {}", get_length(Some(Box::new(node3))));
}

Drag options to blanks, or click blank then click option'

Anext

Bprev

Cchild

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' or other unrelated field names

Inconsistent field names between struct and traversal

The linked list node uses 'next' to point to the next node. The function traverses using 'next'. The example node has 'next' set to None.

Practice

1. What does the function to get the length of a linked list typically do?

easy

A. Counts the number of nodes by traversing from head to end

B. Returns the value of the first node

C. Deletes all nodes in the list

D. Finds the maximum value in the list

Solution

1. Step 1: Understand the linked list traversal

   The function starts at the head node and moves through each node one by one.

2. Step 2: Count each node visited

   For every node visited, the function increments a counter until it reaches the end (null).

3. Final Answer:

   Counts the number of nodes by traversing from head to end -> Option A

4. Quick Check:

   Length = number of nodes visited [OK]

Hint: Traverse and count nodes until null [OK]

Common Mistakes:

• Returning the value of the first node instead of count
• Not handling empty list (null head)
• Trying to count without traversal

2. Which Rust code snippet correctly initializes a linked list node with value 5 and next pointer None?

easy

A. let node = Node::new(5);

B. let node = Node(5, None);

C. let node = Node { value: 5, next: None };

D. let node = Node { value: 5 };

Solution

1. Step 1: Check struct initialization syntax

   Rust structs are initialized with field names and values inside braces.

2. Step 2: Match field names and order

   let node = Node { value: 5, next: None }; uses correct field names and syntax. Tuple syntax requires tuple struct definition. Node::new(5) assumes an impl with new method. Omitting next field causes compile error.

3. Final Answer:

   let node = Node { value: 5, next: None }; -> Option C

4. Quick Check:

   Struct init uses field: value pairs [OK]

Hint: Use struct field names with braces for initialization [OK]

Common Mistakes:

• Using tuple syntax for struct without tuple struct definition
• Missing field names in initialization
• Omitting required fields like next

3. What is the output of this Rust code that calculates the length of a linked list?

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

fn length(head: &Option) -> usize {
    let mut count = 0;
    let mut current = head;
    while let Some(node) = current {
        count += 1;
        current = &node.next;
    }
    count
}

fn main() {
    let node3 = Node { value: 3, next: None };
    let node2 = Node { value: 2, next: Some(Box::new(node3)) };
    let node1 = Node { value: 1, next: Some(Box::new(node2)) };
    let head = Some(Box::new(node1));
    println!("{}", length(&head));
}

medium

A. 1

B. 2

C. 0

D. 3

Solution

1. Step 1: Understand the linked list structure

   There are three nodes linked: node1 -> node2 -> node3 -> None.

2. Step 2: Trace the length function

   The function counts nodes by traversing from head. It visits node1, node2, and node3, incrementing count each time.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   3 nodes counted [OK]

Hint: Count nodes by following next pointers until None [OK]

Common Mistakes:

• Counting only head node
• Not dereferencing Option correctly
• Confusing Option with Box

4. Identify the error in this Rust function to get linked list length:

fn length(head: &Option) -> usize {
    let mut count = 0;
    let mut current = head;
    while current.is_some() {
        count += 1;
        current = current.unwrap().next;
    }
    count
}

medium

A. Using unwrap() on &Option causes a move error

B. Loop condition should be current.is_none()

C. Count should start at 1, not 0

D. Function should return i32, not usize

Solution

1. Step 1: Analyze unwrap usage on reference

   Calling unwrap() on &Option tries to move out of a borrowed reference, which is not allowed.

2. Step 2: Correct way to access next node

   Use pattern matching or while let Some(node) = current to borrow without moving.

3. Final Answer:

   Using unwrap() on &Option causes a move error -> Option A

4. Quick Check:

   unwrap() moves ownership, not allowed on references [OK]

Hint: Use pattern matching, avoid unwrap on references [OK]

Common Mistakes:

• Using unwrap() on borrowed Option causing move error
• Wrong loop condition logic
• Incorrect return type for length

5. Given a linked list where each node contains an integer, how would you modify the length function to count only nodes with even values?

fn length_even(head: &Option) -> usize {
    // Your code here
}

hard

A. Use recursion without checking node values

B. Traverse list, increment count only if node.value % 2 == 0

C. Return length() / 2 assuming half are even

D. Count all nodes, then subtract odd values count

Solution

1. Step 1: Traverse the linked list nodes

   Start from head and move through each node until None.

2. Step 2: Check each node's value for evenness

   Increment count only if node.value % 2 == 0 (even number).

3. Step 3: Return the count of even-valued nodes

   This gives the length of nodes with even values.

4. Final Answer:

   Traverse list, increment count only if node.value % 2 == 0 -> Option B

5. Quick Check:

   Count only even nodes by checking value % 2 == 0 [OK]

Hint: Count nodes while checking even condition [OK]

Common Mistakes:

• Counting all nodes without filtering
• Assuming half nodes are even without check
• Using recursion without base case