> For the complete documentation index, see [llms.txt](https://chunhthanhde.gitbook.io/leetcode-top-interview/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://chunhthanhde.gitbook.io/leetcode-top-interview/topic-8-linked-list/063-remove-nth-node-from-end-of-list.md).

# 19. Remove Nth Node From End of List 🗑️

**Difficulty**: `Medium` - **Tags**: `Linked List`, `Two Pointers`

[LeetCode Problem Link](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)

***

## Problem Statement 📜

Given the head of a linked list, remove the `n`th node from the end of the list and return its head.

***

## Examples 🌟

🔹 **Example 1**:

![](/files/m2yuPj0PA975Q8cMAj6C)

**Input**:

```plaintext
head = [1,2,3,4,5], n = 2
```

**Output**:

```plaintext
[1,2,3,5]
```

***

🔹 **Example 2**:

**Input**:

```plaintext
head = [1], n = 1
```

**Output**:

```plaintext
[]
```

***

🔹 **Example 3**:

**Input**:

```plaintext
head = [1,2], n = 1
```

**Output**:

```plaintext
[1]
```

***

## Constraints ⚙️

* The number of nodes in the list is `sz`.
* `1 <= sz <= 30`
* `0 <= Node.val <= 100`
* `1 <= n <= sz`

***

## Solution 💡

We can solve this problem in one pass by using the **two-pointer technique**. The first pointer moves `n` steps ahead, and then both pointers move together until the first pointer reaches the end. This ensures the second pointer is just before the node to be removed.

***

### Java Solution

```java
class ListNode {
    int val;
    ListNode next;

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

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        // Create a dummy node to simplify edge cases
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode first = dummy;
        ListNode second = dummy;

        // Move first pointer n+1 steps ahead to maintain a gap of n nodes
        for (int i = 0; i <= n; i++) {
            first = first.next;
        }

        // Move both pointers until first reaches the end
        while (first != null) {
            first = first.next;
            second = second.next;
        }

        // Remove the nth node from the end
        second.next = second.next.next;

        return dummy.next;
    }
}
```

***

## Explanation of the Solution

1. **Dummy Node**:
   * A dummy node is used to handle edge cases where the head needs to be removed.
2. **Two-Pointer Technique**:
   * The first pointer moves `n+1` steps ahead, so when it reaches the end, the second pointer is just before the node to be removed.
3. **Node Removal**:
   * Adjust the `next` pointer of the second pointer to skip the node to be removed.

***

## Time Complexity ⏳

* **O(sz)**: The list is traversed once to find and remove the node.

## Space Complexity 💾

* **O(1)**: The solution uses constant space.

***

## Follow-up 🧐

**Could you do this in one pass?**

The above solution achieves the removal in a single pass using the two-pointer approach, maintaining optimal time complexity.

You can find the full solution [here](https://github.com/ChunhThanhDe/Leetcode-Top-Interview/blob/main/Topic%208%20Linked%20List/063%20Remove%20Nth%20Node%20From%20End%20of%20List/Solution.java).