Why Remove Nth Node from End of List in DSA C#?

The Big Idea

What if you could remove the last few items from a list without counting the whole thing?

The Scenario

Imagine you have a long chain of paper clips linked together. You want to remove the 3rd paper clip from the end, but you only have the chain from the start. Counting each clip from the start to find the right one is tiring and easy to mess up.

The Problem

Manually counting each item from the start to find the one near the end is slow and confusing. If the chain is very long, you might lose track or count wrong. Also, you might need to count twice: once to find the length, and again to find the right clip to remove.

The Solution

This method uses two pointers moving through the chain at different speeds. One pointer moves ahead by n steps first, then both move together until the first pointer reaches the end. This way, the second pointer lands exactly on the node before the one to remove, making removal quick and simple.

Before vs After

✗ Before

int length = 0;
Node current = head;
while (current != null) {
  length++;
  current = current.next;
}
int target = length - n;
current = head;
for (int i = 0; i < target - 1; i++) {
  current = current.next;
}
current.next = current.next.next;

✓ After

Node first = head;
Node second = head;
for (int i = 0; i < n; i++) {
  first = first.next;
}
if (first == null) {
  // Remove head node
  head = head.next;
  return;
}
while (first.next != null) {
  first = first.next;
  second = second.next;
}
second.next = second.next.next;

What It Enables

This technique lets you remove nodes from the end of a list in just one pass, saving time and avoiding mistakes.

Real Life Example

Think of a playlist of songs where you want to remove the 2nd last song quickly without counting all songs from the start.

Key Takeaways

Manual counting is slow and error-prone.

Two-pointer method finds the target node in one pass.

Efficient and simple way to remove nodes from the end.

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