Remove Nth Node from End of List in DSA Rust - Build from Scratch

Remove Nth Node from End of List

📖 Scenario: Imagine you have a chain of train cars connected one after another. Sometimes, you want to remove a car counting from the end of the chain. This project helps you learn how to remove the nth car from the end of a linked list of train cars.

🎯 Goal: You will build a Rust program that creates a singly linked list, then removes the nth node from the end of the list, and finally prints the updated list.

📋 What You'll Learn

Create a singly linked list with exact nodes: 1 -> 2 -> 3 -> 4 -> 5 -> None

Add a variable n to specify which node from the end to remove

Implement the logic to remove the nth node from the end of the list

Print the final linked list in the format: 1 -> 2 -> 3 -> 5 -> None

💡 Why This Matters

🌍 Real World

Linked lists are used in many applications like music playlists, undo history in editors, and managing tasks in operating systems. Removing nodes from the end is a common operation.

💼 Career

Understanding linked lists and pointer manipulation is essential for software engineers working with low-level data structures, embedded systems, or performance-critical applications.

1

Create the linked list

Create a singly linked list with nodes containing values 1, 2, 3, 4, and 5 in that order. Use the struct ListNode with fields val: i32 and next: Option>. Create a variable called head that points to the first node.

DSA Rust

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

// Create the linked list with nodes 1 -> 2 -> 3 -> 4 -> 5 -> None
// Your code here

Hint

Start by defining the ListNode struct. Then create nested Box pointers for each node from 5 back to 1.

2

Add the variable n

Add a variable called n and set it to 2. This variable will represent the position from the end of the list of the node to remove.

DSA Rust

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

fn main() {
    let head = Some(Box::new(ListNode {
        val: 1,
        next: Some(Box::new(ListNode {
            val: 2,
            next: Some(Box::new(ListNode {
                val: 3,
                next: Some(Box::new(ListNode {
                    val: 4,
                    next: Some(Box::new(ListNode {
                        val: 5,
                        next: None,
                    })),
                })),
            })),
        })),
    }));

    // Add variable n with value 2
    // Your code here
}

Hint

Just add let n = 2; inside the main function after creating the list.

3

Remove the nth node from the end

Implement the logic to remove the nth node from the end of the linked list. Use two pointers called fast and slow to find the node to remove. Update the list by skipping the target node.

DSA Rust

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

fn main() {
    let head = Some(Box::new(ListNode {
        val: 1,
        next: Some(Box::new(ListNode {
            val: 2,
            next: Some(Box::new(ListNode {
                val: 3,
                next: Some(Box::new(ListNode {
                    val: 4,
                    next: Some(Box::new(ListNode {
                        val: 5,
                        next: None,
                    })),
                })),
            })),
        })),
    }));

    let n = 2;

    // Remove the nth node from the end using fast and slow pointers
    // Your code here
}

Hint

Create a dummy node before the head. Move fast pointer n+1 steps ahead. Then move fast and slow together until fast reaches the end. Remove the node after slow.

4

Print the updated linked list

Print the updated linked list after removing the nth node from the end. Use a loop to print each node's value followed by ->. End with None.

DSA Rust

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

fn main() {
    let mut head = Some(Box::new(ListNode {
        val: 1,
        next: Some(Box::new(ListNode {
            val: 2,
            next: Some(Box::new(ListNode {
                val: 3,
                next: Some(Box::new(ListNode {
                    val: 4,
                    next: Some(Box::new(ListNode {
                        val: 5,
                        next: None,
                    })),
                })),
            })),
        })),
    }));

    let n = 2;

    let mut dummy = Box::new(ListNode { val: 0, next: head });
    let mut fast = &dummy;
    let mut slow = &mut dummy;

    for _ in 0..=n {
        fast = fast.next.as_ref().unwrap();
    }

    while fast.next.is_some() {
        fast = fast.next.as_ref().unwrap();
        slow = slow.next.as_mut().unwrap();
    }

    let next_next = slow.next.as_mut().unwrap().next.take();
    slow.next = next_next;

    head = dummy.next;

    // Print the updated linked list
    // Your code here
}

Hint

Use a while let loop to traverse the list. Print each node's value followed by ->. After the loop, print None.

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