Remove Nth Node from End of List in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output after removing 2nd node from end

What is the output linked list after removing the 2nd node from the end in the given list?

DSA C#

public class ListNode {
    public int val;
    public ListNode next;
    public ListNode(int val=0, ListNode next=null) {
        this.val = val;
        this.next = next;
    }
}

public class Solution {
    public ListNode RemoveNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0, head);
        ListNode first = dummy;
        ListNode second = dummy;
        for (int i = 0; i <= n; i++) {
            first = first.next;
        }
        while (first != null) {
            first = first.next;
            second = second.next;
        }
        second.next = second.next.next;
        return dummy.next;
    }
}

// Input list: 1 -> 2 -> 3 -> 4 -> 5 -> null
// n = 2
// Call: RemoveNthFromEnd(head, 2)

A 2 -> 3 -> 4 -> 5 -> null

B 1 -> 2 -> 4 -> 5 -> null

C 1 -> 2 -> 3 -> 5 -> null

D 1 -> 3 -> 4 -> 5 -> null

💡 Hint

Think about how the two pointers move and which node gets skipped.

The 2nd node from the end is '4'. Removing it results in the list: 1 -> 2 -> 3 -> 5 -> null.

❓ Predict Output

intermediate

Output after removing 1st node from end

What is the output linked list after removing the last node (1st from end) in the given list?

DSA C#

List: 10 -> 20 -> 30 -> 40 -> null
n = 1
Call RemoveNthFromEnd(head, 1)

A 10 -> 20 -> 30 -> null

B 10 -> 20 -> 40 -> null

C 20 -> 30 -> 40 -> null

D 10 -> 30 -> 40 -> null

💡 Hint

Removing the last node means the node with value 40 is removed.

Removing the last node (40) leaves the list: 10 -> 20 -> 30 -> null.

🔧 Debug

advanced

Identify the error in this RemoveNthFromEnd implementation

What error will this code cause when removing the 3rd node from end in a 3-node list?

DSA C#

public ListNode RemoveNthFromEnd(ListNode head, int n) {
    ListNode first = head;
    ListNode second = head;
    for (int i = 0; i < n; i++) {
        first = first.next;
    }
    while (first.next != null) {
        first = first.next;
        second = second.next;
    }
    second.next = second.next.next;
    return head;
}

// List: 1 -> 2 -> 3 -> null
// n = 3
// Call RemoveNthFromEnd(head, 3)

A NullReferenceException at first = first.next when i=2

B NullReferenceException at second.next = second.next.next

C Returns list unchanged

D Removes wrong node (last node instead of 3rd from end)

💡 Hint

Check what happens when first pointer moves n steps ahead without dummy node.

When n equals list length, first becomes null after the loop, so first.next causes NullReferenceException.

🧠 Conceptual

advanced

Why use a dummy node in RemoveNthFromEnd?

What is the main reason to use a dummy node before the head in the RemoveNthFromEnd algorithm?

A To reduce the time complexity from O(n^2) to O(n)

B To simplify removal of the head node when it is the target

C To avoid using two pointers and use only one pointer

D To store the length of the list for easier calculation

💡 Hint

Think about what happens if the node to remove is the first node.

The dummy node allows uniform handling of node removal, including when the head node must be removed, avoiding special cases.

🚀 Application

expert

Resulting list after multiple RemoveNthFromEnd calls

Given the initial list 5 -> 10 -> 15 -> 20 -> 25 -> null, what is the list after these calls? 1) RemoveNthFromEnd(head, 3) 2) RemoveNthFromEnd(head, 1) 3) RemoveNthFromEnd(head, 3)

A 5 -> 15 -> 20 -> 25 -> null

B 5 -> 10 -> 15 -> null

C 10 -> 15 -> 20 -> null

D 10 -> 20 -> null

💡 Hint

Apply each removal step carefully, updating the list each time.

Step 1 removes 15 (3rd from end), list: 5 -> 10 -> 20 -> 25 Step 2 removes 25 (last node), list: 5 -> 10 -> 20 Step 3 removes 5 (3rd from end, which is head), list: 10 -> 20 Final list: 10 -> 20 -> null

Practice

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

easy

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

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

C. Reverse the linked list

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

Solution

1. Step 1: Understand the problem statement

   The problem asks to remove the node that is nth from the end of a singly linked list.

2. Step 2: Clarify what removal means

   Removal means unlinking that node so the list skips it, keeping the rest intact.

3. Final Answer:

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

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 it means reversing the list
• Trying to find value instead of removing node

2. Which of the following C# code snippets correctly declares a singly linked list node class for this problem?

easy

A. public class ListNode { int val; ListNode next; ListNode(int x) { val = x; next = null; } }

B. class Node { int val; Node next; Node(int x) { val = x; next = null; } }

C. public struct ListNode { public int val; public ListNode next; }

D. public class ListNode { public int val; public ListNode next; public ListNode(int x) { val = x; next = null; } }

Solution

1. Step 1: Check for public access modifiers

   public class ListNode { public int val; public ListNode next; public ListNode(int x) { val = x; next = null; } } uses public class and public members, which is correct for external access.

2. Step 2: Verify constructor and member initialization

   public class ListNode { public int val; public ListNode next; public ListNode(int x) { val = x; next = null; } } correctly initializes val and next in constructor; others miss public or use struct incorrectly.

3. Final Answer:

   public class ListNode { public int val; public ListNode next; public ListNode(int x) { val = x; next = null; } } -> Option D

4. Quick Check:

   Proper class with public members = public class ListNode { public int val; public ListNode next; public ListNode(int x) { val = x; next = null; } } [OK]

Hint: Use public class and public members for linked list nodes [OK]

Common Mistakes:

• Missing public keyword causing access errors
• Using struct instead of class for linked list node
• Not initializing next to null in constructor

3. What will be the output linked list after removing the 2nd node from the end in the list: 1 -> 2 -> 3 -> 4 -> 5 -> null?

medium

A. 1 -> 3 -> 4 -> 5 -> null

B. 1 -> 2 -> 4 -> 5 -> null

C. 1 -> 2 -> 3 -> 5 -> null

D. 2 -> 3 -> 4 -> 5 -> null

Solution

1. Step 1: Identify the 2nd node from the end

   The list is 1->2->3->4->5. Counting from end: 5 (1st), 4 (2nd). So node with value 4 is to be removed.

2. Step 2: Remove node with value 4

   Link node 3's next to node 5, skipping node 4. Result: 1->2->3->5->null.

3. Final Answer:

   1 -> 2 -> 3 -> 5 -> null -> Option C

4. Quick Check:

   Remove 2nd from end (4) = 1->2->3->5 [OK]

Hint: Count from end to find node, then skip it [OK]

Common Mistakes:

• Removing the wrong node counting from start
• Removing node with value 3 instead of 4
• Not updating links correctly causing broken list

4. Given the following code snippet to remove the nth node from end, what is the bug?

public ListNode RemoveNthFromEnd(ListNode head, int n) {
  ListNode fast = head;
  ListNode slow = head;
  for (int i = 0; i < n; i++) {
    fast = fast.next;
  }
  while (fast.next != null) {
    fast = fast.next;
    slow = slow.next;
  }
  slow.next = slow.next.next;
  return head;
}

medium

A. Does not handle case when fast becomes null after initial loop

B. Uses slow pointer incorrectly by moving it too early

C. Returns slow instead of head

D. Does not update fast pointer inside the loop

Solution

1. Step 1: Analyze fast pointer movement

   Fast moves n steps ahead. If n equals list length, fast becomes null.

2. Step 2: Check for null fast before accessing fast.next

   Code does not check if fast is null before while loop, causing NullReferenceException.

3. Final Answer:

   Does not handle case when fast becomes null after initial loop -> Option A

4. Quick Check:

   Null fast after n steps = bug [OK]

Hint: Check if fast is null after initial n steps [OK]

Common Mistakes:

• Ignoring null check for fast pointer
• Moving slow pointer before fast is properly positioned
• Returning wrong node as new head

5. You want to remove the 1st node from the end of a linked list with only one node: 10 -> null. What should the resulting list be?

hard

A. 10 -> null

B. null (empty list)

C. Error because list is too short

D. 10 -> 10 -> null

Solution

1. Step 1: Understand removing 1st node from end in single-node list

   Removing last node means list becomes empty.

2. Step 2: Resulting list after removal

   Since only node is removed, list is now null (empty).

3. Final Answer:

   null (empty list) -> Option B

4. Quick Check:

   Remove only node = empty list [OK]

Hint: Removing last node in single-node list empties it [OK]

Common Mistakes:

• Returning original list instead of empty
• Expecting error instead of empty list
• Trying to access next node which does not exist