Pop Using Linked List Node in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to remove the first node from the linked list.

DSA Rust

fn pop(head: &mut Option<Box<Node>>) -> Option<i32> {
    if head.is_none() {
        return None;
    }
    let node = head.take().unwrap();
    *head = node.[1];
    Some(node.value)
}

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

Anext

Bprev

Cvalue

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' causes a compile error.

Assigning 'value' instead of 'next' breaks the list structure.

The 'next' field points to the next node, so assigning head to node.next removes the first node.

2fill in blank

medium

Complete the code to safely pop the first node and return its value.

DSA Rust

fn pop(head: &mut Option<Box<Node>>) -> Option<i32> {
    head.take().map(|node| {
        *head = node.[1];
        node.value
    })
}

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

Aprev

Bhead

Cvalue

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' causes errors.

Returning node.next instead of node.value returns wrong data.

We assign head to node.next to remove the first node and keep the rest of the list.

3fill in blank

hard

Fix the error in the pop function to correctly update the head pointer.

DSA Rust

fn pop(head: &mut Option<Box<Node>>) -> Option<i32> {
    if let Some(node) = head.take() {
        *head = node.[1];
        Some(node.value)
    } else {
        None
    }
}

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

Avalue

Bnext

Chead

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning head to node.value causes type mismatch.

Using node.prev causes compile errors.

The head must be updated to node.next to remove the first node properly.

4fill in blank

hard

Fill both blanks to create a function that pops the first node and returns its value.

DSA Rust

fn pop(head: &mut Option<Box<Node>>) -> Option<i32> {
    if head.is_none() {
        return None;
    }
    let node = head.[1]().unwrap();
    *head = node.[2];
    Some(node.value)
}

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

Atake

Bclone

Cnext

Dunwrap

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'clone' instead of 'take' causes ownership issues.

Assigning head to node.unwrap causes errors.

Use 'take' to remove the node from head and 'next' to update head to the next node.

5fill in blank

hard

Fill all three blanks to implement pop that removes and returns the first node's value.

DSA Rust

fn pop(head: &mut Option<Box<Node>>) -> Option<i32> {
    head.[1]().map(|node| {
        *head = node.[2];
        node.[3]
    })
}

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

Atake

Bnext

Cvalue

Dclone

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'clone' instead of 'take' causes ownership errors.

Returning node.next instead of node.value returns wrong data.

Use 'take' to remove the node, update head to node.next, and return node.value.

Practice

1. What does the pop operation do in a linked list implemented using nodes?

easy

A. Removes and returns the first node from the list

B. Adds a new node at the end of the list

C. Counts the total number of nodes in the list

D. Reverses the order of nodes in the list

Solution

1. Step 1: Understand the pop operation in linked list context

   Pop means removing an element, usually from the start or top of a data structure.

2. Step 2: Identify pop behavior for linked list nodes

   In a linked list, pop removes the first node and returns its value, updating the head pointer.

3. Final Answer:

   Removes and returns the first node from the list -> Option A

4. Quick Check:

   Pop = Remove first node [OK]

Hint: Pop removes the first node in a linked list [OK]

Common Mistakes:

• Thinking pop adds nodes instead of removing
• Confusing pop with counting nodes
• Assuming pop reverses the list

2. Which of the following Rust code snippets correctly pops a node from the head of a singly linked list?

easy

A. let popped = list.head.take(); list.head = popped.as_ref().and_then(|node| node.next.clone());

B. let popped = list.tail.take(); list.tail = popped.as_ref().and_then(|node| node.prev.clone());

C. let popped = list.head.clone(); list.head = popped.unwrap().next.clone();

D. let popped = list.head.unwrap(); list.head = popped.next.unwrap();

Solution

1. Step 1: Identify correct pop syntax for head node

   Pop removes the head node safely using take() to replace head with None temporarily.

2. Step 2: Update head to next node correctly

   After taking head, update head to the next node using as_ref() and and_then to clone next safely.

3. Final Answer:

   let popped = list.head.take(); list.head = popped.as_ref().and_then(|node| node.next.clone()); -> Option A

4. Quick Check:

   Use take() and update head to next [OK]

Hint: Use take() on head and update to next node [OK]

Common Mistakes:

• Trying to pop from tail instead of head
• Using unwrap() unsafely causing panic
• Not updating head after pop

3. Given the Rust code below, what will be printed after popping one node from the linked list?

struct Node { value: i32, next: Option }

let mut head = Some(Box::new(Node { value: 10, next: Some(Box::new(Node { value: 20, next: None })) }));

let popped = head.take();
head = popped.as_ref().and_then(|node| node.next.clone());

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

medium

A. 10 -> null

B. 10 -> 20 -> null

C. null

D. 20 -> null

Solution

1. Step 1: Understand initial list structure

   List has nodes 10 -> 20 -> null, head points to node with value 10.

2. Step 2: Pop removes first node and updates head

   Pop removes node 10, head now points to node 20.

3. Step 3: Print remaining list

   Loop prints node 20 followed by null.

4. Final Answer:

   20 -> null -> Option D

5. Quick Check:

   Pop removes 10, list starts at 20 [OK]

Hint: Pop removes first node; print starts from new head [OK]

Common Mistakes:

• Printing old head node after pop
• Not updating head after pop
• Confusing Option C (empty list) with actual output

4. The following Rust code attempts to pop a node from a linked list but causes a runtime panic. What is the error?

let mut head = None;
let popped = head.unwrap();
head = popped.next;

medium

A. Missing mutable borrow of head

B. Calling unwrap() on None causes panic

C. Cannot assign popped.next to head

D. next field is private and inaccessible

Solution

1. Step 1: Identify unwrap() usage on None

   head is None, calling unwrap() on None causes panic at runtime.

2. Step 2: Understand why panic occurs

   unwrap() expects Some value; None triggers panic error.

3. Final Answer:

   Calling unwrap() on None causes panic -> Option B

4. Quick Check:

   unwrap() on None = panic [OK]

Hint: Never unwrap None; check before unwrap [OK]

Common Mistakes:

• Assuming unwrap() returns None safely
• Ignoring that head is None initially
• Thinking mutable borrow causes panic

5. You have a linked list implemented with nodes in Rust. You want to pop all nodes one by one and collect their values into a vector. Which approach correctly achieves this?

hard

A. Use a for loop over head and push values, then set head to None

B. Use recursion: call pop on head until head is None, collecting values in vector

C. Use a loop: while let Some(node) = head.take() { push node.value; head = node.next; }

D. Clone the entire list, then pop nodes from the clone to preserve original

Solution

1. Step 1: Use while let loop to pop nodes safely

   Loop uses head.take() to remove head node safely each iteration.

2. Step 2: Push node values and update head

   Inside loop, push node.value to vector and update head to node.next for next iteration.

3. Step 3: Avoid recursion or incorrect iteration

   Recursion can cause stack overflow; for loop over Option is invalid; cloning wastes memory.

4. Final Answer:

   Use a loop: while let Some(node) = head.take() { push node.value; head = node.next; } -> Option C

5. Quick Check:

   Loop with take() and update head [OK]

Hint: Use while let with take() to pop all nodes [OK]

Common Mistakes:

• Using recursion causing stack overflow
• Trying to iterate Option with for loop
• Cloning list unnecessarily wasting memory