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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output after removing 2nd node from end

Given the linked list 1 -> 2 -> 3 -> 4 -> 5 -> null, what is the output after removing the 2nd node from the end?

DSA Java

class ListNode {
    int val;
    ListNode next;
    ListNode(int val) { this.val = val; }
}

public class Solution {
    public static ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0);
        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;
        return dummy.next;
    }

    public static void main(String[] args) {
        ListNode head = new ListNode(1);
        head.next = new ListNode(2);
        head.next.next = new ListNode(3);
        head.next.next.next = new ListNode(4);
        head.next.next.next.next = new ListNode(5);
        ListNode result = removeNthFromEnd(head, 2);
        while (result != null) {
            System.out.print(result.val + " -> ");
            result = result.next;
        }
        System.out.print("null");
    }
}

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

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

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

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

💡 Hint

Remember the 2nd node from the end is the 4th node from the start in a 5-node list.

Removing the 2nd node from the end removes the node with value 4, so the list becomes 1 -> 2 -> 3 -> 5 -> null.

❓ Predict Output

intermediate

Output after removing 1st node from end

Given the linked list 10 -> 20 -> 30 -> null, what is the output after removing the 1st node from the end?

DSA Java

class ListNode {
    int val;
    ListNode next;
    ListNode(int val) { this.val = val; }
}

public class Solution {
    public static ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0);
        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;
        return dummy.next;
    }

    public static void main(String[] args) {
        ListNode head = new ListNode(10);
        head.next = new ListNode(20);
        head.next.next = new ListNode(30);
        ListNode result = removeNthFromEnd(head, 1);
        while (result != null) {
            System.out.print(result.val + " -> ");
            result = result.next;
        }
        System.out.print("null");
    }
}

A 20 -> 30 -> null

B 10 -> 30 -> null

C 10 -> 20 -> 30 -> null

D 10 -> 20 -> null

💡 Hint

Removing the last node means the node with value 30 is removed.

The last node (value 30) is removed, so the list becomes 10 -> 20 -> null.

🔧 Debug

advanced

Identify the error in removing nth node from end

What error does the following code produce when removing the 3rd node from the end of list 1 -> 2 -> 3 -> 4 -> 5 -> null?

DSA Java

class ListNode {
    int val;
    ListNode next;
    ListNode(int val) { this.val = val; }
}

public class Solution {
    public static ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode first = dummy;
        ListNode second = dummy;
        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 dummy.next;
    }
}

A NullPointerException

B IndexOutOfBoundsException

C No error, works correctly

D Compilation error

💡 Hint

Check how many times the first pointer moves ahead before the loop.

The for loop moves first pointer n times instead of n+1, so when n equals list length, first becomes null and first.next causes NullPointerException.

🧠 Conceptual

advanced

Why use a dummy node in removing nth node from end?

Why is a dummy node used at the start of the linked list when removing the nth node from the end?

A To simplify edge cases like removing the head node

B To increase the speed of traversal

C To store the length of the list

D To avoid using two pointers

💡 Hint

Think about what happens if the head node itself needs to be removed.

Dummy node helps handle cases where the node to remove is the head, avoiding null pointer errors and simplifying code.

🚀 Application

expert

Resulting list after multiple removals

Starting with the list 5 -> 10 -> 15 -> 20 -> 25 -> null, remove the 3rd node from the end twice in a row. What is the final list?

DSA Java

class ListNode {
    int val;
    ListNode next;
    ListNode(int val) { this.val = val; }
}

public class Solution {
    public static ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0);
        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;
        return dummy.next;
    }

    public static void main(String[] args) {
        ListNode head = new ListNode(5);
        head.next = new ListNode(10);
        head.next.next = new ListNode(15);
        head.next.next.next = new ListNode(20);
        head.next.next.next.next = new ListNode(25);

        head = removeNthFromEnd(head, 3);
        head = removeNthFromEnd(head, 3);

        ListNode curr = head;
        while (curr != null) {
            System.out.print(curr.val + " -> ");
            curr = curr.next;
        }
        System.out.print("null");
    }
}

A 5 -> 10 -> 20 -> null

B 5 -> 10 -> 25 -> null

C 10 -> 15 -> 25 -> null

D 5 -> 15 -> 25 -> null

💡 Hint

Remove the 3rd node from end twice and track the list after each removal.

First removal removes node with value 15 (3rd from end). List becomes 5 -> 10 -> 20 -> 25. Second removal removes node with value 20 (3rd from end now). Final list is 5 -> 10 -> 25 -> null.

Practice

1. What does the problem "Remove Nth Node from End of List" ask you to do with a linked list?

easy

A. Reverse the entire linked list

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

C. Find the value of the Nth node from the start

D. Remove the node that is Nth from the end of the list

Solution

1. Step 1: Understand the problem statement

   The problem requires removing the node that is Nth from the end of the linked list, not from the start.

2. Step 2: Clarify the operation on the list

   We do not add or reverse nodes; we specifically remove one node counting from the end.

3. Final Answer:

   Remove the node that is Nth from the end of the list -> Option D

4. Quick Check:

   Remove Nth from end = Remove Nth from end [OK]

Hint: Focus on 'Nth from end' means counting backwards [OK]

Common Mistakes:

• Confusing Nth from end with Nth from start
• Thinking the problem asks to add or reverse nodes
• Ignoring the direction of counting nodes

2. Which of the following Java method signatures correctly represents a function to remove the Nth node from the end of a singly linked list?

easy

A. public ListNode removeNthFromEnd(ListNode head, int n)

B. public void removeNthFromEnd(int n, ListNode head)

C. public ListNode removeNode(ListNode head, int index)

D. public ListNode deleteNthNode(ListNode node, int n)

Solution

1. Step 1: Check method parameters and return type

   The method should take the head of the list and an integer n, and return the new head after removal.

2. Step 2: Match signature with problem requirements

   public ListNode removeNthFromEnd(ListNode head, int n) matches the expected signature: it takes (ListNode head, int n) and returns ListNode.

3. Final Answer:

   public ListNode removeNthFromEnd(ListNode head, int n) -> Option A

4. Quick Check:

   Correct signature = public ListNode removeNthFromEnd(ListNode head, int n) [OK]

Hint: Method must return updated head and accept head, n as parameters [OK]

Common Mistakes:

• Wrong parameter order
• Void return type instead of ListNode
• Using incorrect parameter types

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 -> 2 -> 4 -> 5 -> null

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

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

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

Solution

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

   Counting from the end: 5 (1st), 4 (2nd). So node with value 4 is to be removed.

2. Step 2: Remove node with value 4 and link previous node to next

   After removal, list becomes 1 -> 2 -> 3 -> 5 -> null.

3. Final Answer:

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

4. Quick Check:

   Remove 2nd from end = Remove node 4 [OK]

Hint: Count nodes from end to find which to remove [OK]

Common Mistakes:

• Removing wrong node counting from start
• Off-by-one errors in counting
• Not updating links correctly after removal

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

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 != null) {
        fast = fast.next;
        slow = slow.next;
    }
    slow.next = slow.next.next;
    return head;
}

medium

A. The for loop should run from i = 0 to i < n, not i <= n

B. The slow pointer should start at null, not head

C. The method should return slow, not head

D. The while loop condition should be fast.next != null

Solution

1. Step 1: Analyze the for loop for moving fast pointer

   The loop runs i from 0 to n inclusive (i <= n), moving fast n+1 steps, which is one step too many and causes NullPointerException.

2. Step 2: Correct the loop to run i < n to move fast exactly n steps ahead

   This ensures fast is n nodes ahead of slow, allowing correct removal of the target node.

3. Final Answer:

   The for loop should run from i = 0 to i < n, not i <= n -> Option A

4. Quick Check:

   Loop count off by one = The for loop should run from i = 0 to i < n, not i <= n [OK]

Hint: Check loop bounds carefully for pointer steps [OK]

Common Mistakes:

• Using i <= n instead of i < n in loop
• Starting slow pointer incorrectly
• Returning wrong node after removal

5. You have a linked list: 10 -> 20 -> 30 -> 40 -> 50 -> null. You want to remove the 5th node from the end. What will be the resulting list after removal?

hard

A. 30 -> 40 -> 50 -> null

B. 10 -> 20 -> 30 -> 40 -> null

C. 20 -> 30 -> 40 -> 50 -> null

D. 10 -> 20 -> 30 -> 40 -> 50 -> null

Solution

1. Step 1: Identify the 5th node from the end

   Since the list has 5 nodes, the 5th from the end is the first node with value 10.

2. Step 2: Remove the first node and update head

   Removing node 10 means the new head is node 20, so list becomes 20 -> 30 -> 40 -> 50 -> null.

3. Final Answer:

   20 -> 30 -> 40 -> 50 -> null -> Option C

4. Quick Check:

   Remove first node = 20 -> 30 -> 40 -> 50 -> null [OK]

Hint: If N equals list length, remove head node [OK]

Common Mistakes:

• Not updating head when first node is removed
• Removing wrong node counting from end
• Confusing list length with N