Remove Nth Node from End of List in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a dummy node pointing to the head.

DSA Rust

let mut dummy = ListNode { val: 0, next: Some(Box::new([1])) };

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

Ahead

Bdummy

CNone

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Using None instead of head causes the list to be empty.

Using dummy or tail is incorrect because they are not the head.

We create a dummy node whose next points to the head of the list to simplify edge cases.

2fill in blank

medium

Complete the code to move the fast pointer n steps ahead.

DSA Rust

for _ in 0..[1] {
    fast = fast.unwrap().next.as_ref();
}

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

An

B1

Cn-1

Dn+1

Attempts:

3 left

💡 Hint

Common Mistakes

Moving fast n+1 steps causes off-by-one errors.

Moving fast n-1 steps causes incorrect gap.

We move the fast pointer exactly n steps ahead to maintain the gap.

3fill in blank

hard

Fix the error in the code to correctly remove the nth node from the end.

DSA Rust

slow.as_mut().unwrap().next = slow.as_mut().unwrap().next.as_mut().unwrap().[1].take();

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

Aprev

Bval

Chead

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' or 'val' causes compilation errors.

Using 'head' is not a field of the node.

To remove the node, we link slow's next to the node after the next node, which is slow.next.next.

4fill in blank

hard

Fill both blanks to move fast and slow pointers until fast reaches the end.

DSA Rust

while fast.is_some() {
    fast = fast.unwrap().next.as_ref();
    slow = slow.[1]().unwrap();
    slow = slow.[2]();
}

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

Aas_mut

Bas_ref

Cunwrap

Dtake

Attempts:

3 left

💡 Hint

Common Mistakes

Using as_ref() gives immutable reference, so we cannot modify.

Using take() removes the value, which is not intended here.

We use as_mut() to get a mutable reference and unwrap() to access the inner value for slow pointer movement.

5fill in blank

hard

Fill all three blanks to return the modified list after removal.

DSA Rust

return dummy.[1].[2]();

// Where dummy is ListNode and next is Option<Box<ListNode>>

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

Anext

Btake

Cunwrap

Das_ref

Attempts:

3 left

💡 Hint

Common Mistakes

Using as_ref() returns a reference, not ownership.

Not using take() causes borrowing errors.

We return dummy.next.take().unwrap() to get ownership of the new head after removal.

Practice

1. What does the problem Remove Nth Node from End of List ask you to do?

easy

A. Reverse the linked list up to the nth node

B. Add a new node at the nth position from the start

C. Delete the node that is the nth from the end in a linked list

D. Find the value of the nth node from the start

Solution

1. Step 1: Understand the problem statement

   The problem requires removing the node that is nth from the end in a singly linked list.

2. Step 2: Clarify the operation

   We do not add or reverse nodes; we specifically delete the nth node counting from the list's end.

3. Final Answer:

   Delete the node that is the nth from the end in a linked list -> Option C

4. Quick Check:

   Remove nth from end = Delete nth from end [OK]

Hint: Focus on 'remove nth from end' means delete that node [OK]

Common Mistakes:

• Confusing nth from end with nth from start
• Thinking the problem is about reversing the list
• Assuming insertion instead of deletion

2. Which Rust code snippet correctly declares a singly linked list node struct for this problem?

easy

A. struct ListNode { val: i32, next: &ListNode }

B. struct ListNode { val: i32, next: Option }

C. struct ListNode { val: i32, next: Box<ListNode> }

D. struct ListNode { val: i32, next: Option<&ListNode> }

Solution

1. Step 1: Recall Rust ownership for linked lists

   Rust uses Option to allow nullable next pointers with ownership.

2. Step 2: Check each option

   struct ListNode { val: i32, next: Option } uses Option which is correct. Options A and D use references without ownership, which is unsafe. struct ListNode { val: i32, next: Box<ListNode> } lacks Option, so no null end.

3. Final Answer:

   struct ListNode { val: i32, next: Option } -> Option B

4. Quick Check:

   Rust linked list node = Option [OK]

Hint: Use Option for nullable owned pointers in Rust [OK]

Common Mistakes:

• Using references (&) instead of Box for next
• Omitting Option causing no null end
• Forgetting ownership rules in Rust

3. Given the Rust code below, what is the linked list output after removing the 2nd node from the end?

let mut list = Some(Box::new(ListNode { val: 1, next: Some(Box::new(ListNode { val: 2, next: Some(Box::new(ListNode { val: 3, next: None })) })) }));
// Remove 2nd from end
let result = remove_nth_from_end(list, 2);
// Print list values

medium

A. 1 -> 2 -> 3 -> null

B. 1 -> 2 -> null

C. 2 -> 3 -> null

D. 1 -> 3 -> null

Solution

1. Step 1: Identify the list and node to remove

   The list is 1 -> 2 -> 3. Removing 2nd from end means removing node with value 2.

2. Step 2: Remove the node and update links

   After removal, list becomes 1 -> 3 -> null.

3. Final Answer:

   1 -> 3 -> null -> Option D

4. Quick Check:

   Remove 2nd from end removes '2' [OK]

Hint: Count from end, remove that node, link neighbors [OK]

Common Mistakes:

• Removing the wrong node counting from start
• Not updating next pointers correctly
• Confusing 2nd from end with 2nd from start

4. The following Rust function aims to remove the nth node from the end but has a bug. What is the bug?

fn remove_nth_from_end(head: Option, n: i32) -> Option {
    let mut dummy = Box::new(ListNode { val: 0, next: head });
    let mut first = &dummy;
    let mut second = &mut dummy;
    for _ in 0..n {
        first = first.next.as_ref().unwrap();
    }
    while first.next.is_some() {
        first = first.next.as_ref().unwrap();
        second = second.next.as_mut().unwrap();
    }
    second.next = second.next.as_mut().unwrap().next.take();
    dummy.next
}

medium

A. Using unwrap() without checking may panic if list is short

B. second is immutable, cannot modify next pointer

C. first is an immutable reference, cannot advance correctly

D. dummy.next is returned before modification

Solution

1. Step 1: Analyze unwrap usage

   The code uses unwrap() on next without checking if next exists, which can panic if n is larger than list length.

2. Step 2: Confirm other references are valid

   first is immutable reference but used only for reading, second is mutable for modification, dummy.next returned after modification.

3. Final Answer:

   Using unwrap() without checking may panic if list is short -> Option A

4. Quick Check:

   unwrap() without safety check causes panic [OK]

Hint: Check unwrap() calls for possible panics [OK]

Common Mistakes:

• Confusing mutable and immutable references
• Ignoring unwrap panic risks
• Assuming dummy.next is returned too early

5. You want to remove the nth node from the end of a linked list in Rust. Which approach correctly handles the edge case when the node to remove is the head node?

hard

A. Use a dummy node before head and adjust dummy.next after removal

B. Directly remove head without dummy node by shifting head pointer

C. Traverse twice: first to count length, second to remove node without dummy

D. Convert list to vector, remove element, then rebuild list

Solution

1. Step 1: Understand the edge case

   If the node to remove is the head, removing it directly is tricky because head pointer must change.

2. Step 2: Use dummy node technique

   Adding a dummy node before head simplifies removal by always having a previous node to update, including when head is removed.

3. Final Answer:

   Use a dummy node before head and adjust dummy.next after removal -> Option A

4. Quick Check:

   Dummy node handles head removal edge case [OK]

Hint: Add dummy node to simplify head removal [OK]

Common Mistakes:

• Not handling head removal separately
• Forgetting to update head pointer
• Using inefficient vector conversion