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

Remove Nth Node from End of List

📖 Scenario: You are working on a music playlist app. The playlist is a linked list where each node holds a song's name. Sometimes, users want to remove a song counting from the end of the playlist. Your task is to remove the nth song from the end of the playlist.

🎯 Goal: Build a program that creates a linked list of songs, then removes the nth node from the end of the list, and finally prints the updated playlist.

📋 What You'll Learn

Create a singly linked list with exact song names in order

Create an integer variable n representing the position from the end

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

Print the updated linked list showing song names separated by ' -> ' and ending with 'null'

💡 Why This Matters

🌍 Real World

Removing the nth node from the end of a linked list is useful in applications like music playlists, task queues, or browsing history where you want to delete items counting from the end.

💼 Career

Understanding linked list manipulation and two-pointer techniques is important for software engineering roles that involve data structure optimization and algorithmic problem solving.

1

Create the linked list of songs

Create a singly linked list with the class ListNode where each node has a string val and a ListNode next. Then create the linked list with these exact songs in order: "SongA", "SongB", "SongC", "SongD", "SongE". Store the head node in a variable called head.

DSA C#

// Define ListNode class and create linked list with songs SongA -> SongB -> SongC -> SongD -> SongE
// Your code here

Hint

Define the ListNode class first. Then link nodes one by one to form the list.

2

Set the position n from the end

Create an integer variable called n and set it to 2. This means you want to remove the 2nd song from the end of the playlist.

DSA C#

public class ListNode {
    public string val;
    public ListNode next;
    public ListNode(string val= "", ListNode next=null) {
        this.val = val;
        this.next = next;
    }
}

ListNode head = new ListNode("SongA");
head.next = new ListNode("SongB");
head.next.next = new ListNode("SongC");
head.next.next.next = new ListNode("SongD");
head.next.next.next.next = new ListNode("SongE");

// Create an integer variable n and set it to 2
// Your code here

Hint

Just create a variable n and assign it the value 2.

3

Remove the nth node from the end

Write code to remove the nth node from the end of the linked list. Use two pointers: first and second, both starting at a dummy node before head. Move first ahead by n + 1 steps, then move both pointers until first reaches the end. Then remove the node after second. Update head to dummy.next.

DSA C#

public class ListNode {
    public string val;
    public ListNode next;
    public ListNode(string val= "", ListNode next=null) {
        this.val = val;
        this.next = next;
    }
}

ListNode head = new ListNode("SongA");
head.next = new ListNode("SongB");
head.next.next = new ListNode("SongC");
head.next.next.next = new ListNode("SongD");
head.next.next.next.next = new ListNode("SongE");

int n = 2;

// Remove the nth node from the end using two pointers
// Your code here

Hint

Use a dummy node before head to handle edge cases. Move first pointer n+1 steps ahead, then move both pointers until first reaches the end. Remove the node after second.

4

Print the updated playlist

Write code to print the updated linked list starting from head. Print each song's name followed by " -> ". After the last song, print "null".

DSA C#

public class ListNode {
    public string val;
    public ListNode next;
    public ListNode(string val= "", ListNode next=null) {
        this.val = val;
        this.next = next;
    }
}

ListNode head = new ListNode("SongA");
head.next = new ListNode("SongB");
head.next.next = new ListNode("SongC");
head.next.next.next = new ListNode("SongD");
head.next.next.next.next = new ListNode("SongE");

int n = 2;

ListNode dummy = new ListNode("");
dummy.next = 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;
head = dummy.next;

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

Hint

Use a loop to go through each node starting from head. Print the song name followed by ' -> '. After the loop, print '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