Why Remove Nth Node from End of List in DSA Java?

The Big Idea

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

The Scenario

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

The Problem

Counting from the start every time to find the clip near the end is slow and confusing. If the chain is very long, you might lose track or count wrong. You might also accidentally remove the wrong clip, breaking the chain.

The Solution

This method lets you find and remove the clip from the end in just one pass through the chain. It uses two pointers moving at different speeds to find the exact clip without counting the whole chain first.

Before vs After

✗ Before

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

✓ After

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

What It Enables

You can quickly and safely remove any node from the end of a linked list in just one pass, no matter how long the list is.

Real Life Example

Think of a playlist of songs where you want to remove the 2nd last song without counting all songs from the start. This method helps you do that efficiently.

Key Takeaways

Manual counting is slow and error-prone for long lists.

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

Efficient removal from end without full list traversal.

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