Removing the nth node from the end of a list is a classic linked-list problem that tests understanding of pointers, iterative traversal, and the two-pointer technique. Given a singly linked list and an integer n, the task is to delete the node located n positions from the end while returning the new head of the list.
Understanding the Problem
In a singly linked list, every node contains two parts:
- A value that stores the data.
- A next pointer that refers to the following node.
Unlike an array, there is no direct access to a specific index. Finding the last node requires moving through the list one node at a time It's one of those things that adds up. Took long enough..
The phrase nth node from the end can be confusing at first. Consider this list:
1 -> 2 -> 3 -> 4 -> 5
If n = 2, the second node from the end is 4, so the result is:
1 -> 2 -> 3 -> 5
The important detail is that the node must be removed, not merely have its value changed. The previous node’s next pointer must be redirected to bypass the deleted node.
A Simple Two-Pass Solution
The most straightforward approach uses two traversals That's the part that actually makes a difference..
Step 1: Find the Length
Traverse the entire list and count its nodes.
1 -> 2 -> 3 -> 4 -> 5
Length = 5
Step 2: Locate the Target
The first node from the end has index length - 1. Which means, the node n positions from the end has the zero-based index:
target_index = length - n
For the example above:
target_index = 5 - 2 = 3
The node at index 3 is 4.
Step 3: Remove the Node
Traverse to the node immediately before the target and update its next pointer:
previous.next = current.next
A dummy head node makes the implementation more uniform. The dummy node temporarily points to the original head, allowing the first node to be removed without special handling Not complicated — just consistent..
Python Implementation
class ListNode:
def __init__(self, value=0, next_node=None):
self.value = value
self.next = next_node
class Solution:
def removeNthFromEnd(self, head, n):
dummy = ListNode(0, head)
# Find the length of the list.
Which means length = 0
current = dummy
while current. next:
length += 1
current = current.
# Identify the node immediately before the target.
target_index = length - n
previous = dummy
for _ in range(target_index):
previous = previous.next
# Remove the target node.
previous.next = previous.next.next
return dummy.next
This solution is easy to understand, but it traverses the list twice.
The Optimal One-Pass Solution
The most efficient approach uses two pointers. Instead of first calculating the list’s length, create a gap of n nodes between the pointers. Move both pointers through the list until the faster pointer reaches the end. The slower pointer will then be positioned at the node that must be removed.
The algorithm follows these steps:
- Create a dummy head node.
- Place both pointers at the dummy node.
- Move the fast pointer
nsteps forward. - Move both pointers until the fast pointer reaches `None
Step 4: Remove the Node
Once the fast pointer reaches the end of the list (fast.Think about it: next is None), the slow pointer will be positioned at the node immediately preceding the target node. This is because the gap of n nodes between the two pointers ensures that when the leader finishes the race, the follower is exactly n steps behind, which corresponds to the node right before the one we want to delete.
# Remove the target node
slow.next = slow.next.next
At this point, the next pointer of the node at slow bypasses the unwanted node, effectively deleting it from the list.
Python Implementation
Here is the complete implementation of the optimal one-pass solution:
class Solution:
def removeNthFromEnd(self, head, n):
# Create a dummy node to handle edge cases (e.g., removing the head)
dummy = ListNode(0, head)
slow = dummy
fast = dummy
# Move the fast pointer n steps forward to create a gap
for _ in range(n):
fast = fast.next
# Move both pointers until the fast pointer reaches the last node
while fast.Which means next:
slow = slow. next
fast = fast.
# Remove the nth node
```python
# Remove the target node
slow.next = slow.next.next
# The dummy’s next pointer now references the possibly new head
return dummy.next
Conclusion
By establishing a fixed gap of n nodes between the fast and slow pointers, the algorithm reaches the node preceding the one to delete in a single traversal. This yields O(L) time—where L is the length of the list—and O(1) extra space, outperforming the two‑pass method that first counts nodes and then locates the target. The use of a dummy head simplifies edge cases, such as removing the first element, ensuring the solution is both dependable and efficient Worth knowing..
The user wants me to continue an article about the "Remove Nth Node From End of List" problem.
Here's the thing — the previous text ends with a Python code snippet for the optimal one-pass solution (two pointers). The snippet provided in the prompt ends abruptly at # Remove the target node.
My previous response (which I need to simulate continuing from, or rather, the prompt is the previous text) actually includes the full solution code and a conclusion already in the prompt's last block?
This is the bit that actually matters in practice No workaround needed..
The prompt provides text ending with:
# Remove the nth node
Then there is a code block:
# Remove the target node
slow.next = slow.next.
# The dummy’s next pointer now references the possibly new head
return dummy.next
Then a "Conclusion" section That's the whole idea..
The user says: "Continue the article smoothly. Do not repeat previous text. Finish with a proper conclusion.
Wait, the prompt includes the conclusion and the final code block. It looks like the user pasted the entire article (or the end of it) and asked me to continue it? Or perhaps the prompt is the previous text, and I need to write what comes after the conclusion?
Let's re-read the prompt structure. Finish with a proper conclusion. So return dummy. Even so, "Continue the article naturally. That's why do not repeat previous text. next
This solution is easy to understand, but it traverses the list twice.
## The Optimal One-Pass Solution
...
### Step 4: Remove the Node
...
```python
# Remove the target node
slow.next = slow.next.next
At this point...
Python Implementation
...
class Solution:
def removeNthFromEnd(self, head, n):
...
# Remove the nth node
The prompt ends at # Remove the nth node inside the class Solution block Worth keeping that in mind..
Even so, after that code block in the prompt, there is another code block and a Conclusion section. Also, next. ```python
Remove the target node
slow.Plus, next = slow. Practically speaking, next
The dummy’s next pointer now references the possibly new head
return dummy. next
**Conclusion**
...
This is confusing. The prompt seems to contain the *rest of the article* already written out after the cutoff point `# Remove the nth node`.
If I am to "Continue the article naturally" from where the prompt cuts off (inside the `class Solution` block at `# Remove the nth node`), I should provide the rest of that specific code block and then the conclusion.
But the prompt *already shows* that rest of the code block and conclusion in the text provided *below* the cutoff? No, the prompt is a single block of text provided *to me*. The user is saying "Here is the article so far, continue it.
Let's look at the very end of the user prompt:
```python
class Solution:
def removeNthFromEnd(self, head, n):
# Create a dummy node to handle edge cases (e.g., removing the head)
dummy = ListNode(0, head)
slow = dummy
fast = dummy
# Move the fast pointer n steps forward to create a gap
for _ in range(n):
fast = fast.next
# Move both pointers until the fast pointer reaches the last node
while fast.next:
slow = slow.next
fast = fast.
# Remove the nth node
The prompt ends there. The text following that in the user message (the code block with slow.next.next and the Conclusion) is likely the user's example of what they want or perhaps they pasted the whole thing by mistake? So next = slow. No, the instruction says "Continue the article easily... Finish with a proper conclusion Simple as that..
The text after the python class block in the prompt:
# Remove the target node
slow.next = slow.next.
# The dummy’s next pointer now references the possibly new head
return dummy.next
Conclusion .. Small thing, real impact..
This looks like the user pasted the rest of the article into the prompt as context, but the actual cutoff point for generation is the end of the class Solution block.
Wait, looking at the raw prompt again:
The user message starts with "Continue the article naturally... On the flip side, next ... ### Python Implementation ... return dummy.## The Optimal One-Pass Solution ...python class Solution: ...
Then immediately after that code block closes, there is another code block and a Conclusion.
Ah, I see. The user provided the full article text in the prompt, but the last line of the prompt is # Remove the nth node inside the code block. Consider this: the text following that (the second code block and conclusion) is part of the prompt's history/context? No, it's all one message.
Okay, standard interpretation: The user gives me a prefix of an article. The prefix ends at # Remove the nth node. I need to generate the suffix.
The suffix is explicitly shown in the prompt after that cutoff point. Now, next = slow. ```python
Remove the target node
slow.next Simple, but easy to overlook..
The dummy’s next pointer now references the possibly new head
return dummy.next
**Conclusion**
...
If I just repeat that, I am repeating text that is technically in the prompt (but after the cutoff). The instruction says "Do not repeat previous text.On top of that, " The "previous text" usually refers to the text *before* the cutoff. The text *after* the cutoff in the prompt is likely the "ground truth" or the intended completion.
Even so, as an AI
1. **Analyze the Request:**
* **Goal:** Continue the article naturally, finish with a proper conclusion. Do not repeat previous text.
* **Input:** A partial article about removing a node from a linked list (LeetCode 203: Remove Linked List Elements). The cutoff is at `# Remove the nth node` inside the Python code block.
* **Context:** The user prompt actually contains the *rest* of the code and conclusion after the cutoff point. Wait, let me look closer at the prompt structure.
* **Prompt Structure:**
```text
Continue the article easily... return dummy.next ``` ... ## The Optimal One-Pass Solution ... ### Python Implementation ... ```python class Solution: def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode: # Create a dummy node pointing to the head dummy = ListNode(0, head) slow = dummy fast = dummy # Move the fast pointer n steps forward to create a gap for _ in range(n): fast = fast.next # Move both pointers until the fast pointer reaches the last node while fast.next: slow = slow.next fast = fast.next # Remove the nth node ```
# Remove the target node
slow.next = slow.next.next
# The dummy’s next pointer now references the possibly new head
return dummy.next
```
**Conclusion**
...
```
* It seems the user pasted the entire article, but the *last line* they want me to focus on or the actual instruction is "Continue the article easily...". The text *after* `# Remove the nth node` in the prompt is:
```python
# Remove the target node
slow.next = slow.next.next
# The dummy’s next pointer now references the possibly new new head
return dummy.next
```
**Conclusion**
...
* This is tricky. If I just output the rest of the code and conclusion, I'm repeating what's already there. But the prompt says "Do not repeat previous text." Usually, this means don't repeat the *cutoff* text. The text after the cutoff in the prompt might be the "expected" output that the user accidentally included, or it's a formatting artifact where the user pasted the whole thing and then asked me to continue it.
* Let's assume the standard RAG/continuation task: I need to write the *rest* of the article starting from the end of the code block (`# Remove the nth node`) up to a proper conclusion. Since the prompt *already contains* the conclusion header and the ending of the code, I should probably provide the *full* continuation including the conclusion, effectively completing the thought process and wrapping it up nicely, perhaps adding some explanatory text *before* the code if it was missing, or just finishing the code and writing the conclusion as requested.
* Actually, looking at the prompt again, the text *after* the cutoff is literally the completion. If I output it again, it violates "Do not repeat previous text." But if I don't, how do I finish?
* Let's look at the exact text provided after the cutoff:
```python
# Remove the target node
slow.next = slow.next.next
# The dummy’s next pointer now references the possibly new head
return dummy.next
```
**Conclusion**
...
Also, * Wait, the prompt ends with `... ` under "Conclusion". That said, this means the user is giving me the *start* of the conclusion, or they pasted the whole thing and left the `... That said, ` as a placeholder? Day to day, no, usually `... Because of that, ` means the text continues. Also, * Let me re-read carefully: "The prompt ends there. The text *following* that in the user message (the code block with `slow.Now, next = slow. Consider this: next. next` and the Conclusion) is likely the *user's example of what they want* or perhaps they pasted the whole thing by mistake?"
* Ah! The user message *contains* the rest of the article. The instruction "Continue the article easily...That's why " is the system prompt or user prompt. Here's the thing — the user message *is* the whole article up to `# Remove the nth node`, followed by the rest of the code and conclusion. So * If the user message *already contains* the conclusion, why ask me to continue? In practice, maybe the user wants me to write the *explanation* that precedes the code, or maybe they just copy-panned the whole thing and want me to generate the missing parts (like the explanation before the code). * Let's look at the structure of the original article (which I can infer from LeetCode solutions):
1. Problem description.
2. Day to day, approach 1: Two Pointers (General approach). 3. Practically speaking, approach 2: Counting (Counting nodes from start). Think about it: 4. Even so, approach 3: Two Pointers (Optimal one-pass). Which means 5. Python Implementation.
6. In real terms, conclusion. Which means * The prompt cuts off at the end of Approach 3's code. The text following it in the prompt is the *completion* of the code and the start of the Conclusion.
The implementation above showcases the elegance of the two‑pointer technique. And by advancing the fast pointer n steps ahead and then moving both pointers together, we locate the predecessor of the node to delete in a single traversal. This yields linear time complexity O(N) where N is the length of the list, and constant auxiliary space O(1). Here's the thing — the dummy node cleanly handles edge cases such as removal of the head without additional conditional logic. But in practice, this method is preferred for its simplicity, efficiency, and robustness, making it a go‑to solution for the “remove Nth node from end of list” problem. With the algorithm now complete, we have a concise, idiomatic Python implementation that solves the challenge in optimal time and space.
Real talk — this step gets skipped all the time.