Practice/Meta/Leetcode 19. Remove Nth Node From End of List

Leetcode 19. Remove Nth Node From End of List

CodingMust

Problem

Given the head of a singly linked list, remove the node that is k positions from the end of the list and return the modified list's head.

The position counting starts at 1, where position 1 from the end is the last node (tail), position 2 from the end is the second-to-last node, and so on.

Requirements

1. Remove exactly one node from the linked list
2. The node to remove is specified by its position from the end (1-indexed)
3. Return the head of the modified linked list
4. If the list becomes empty after removal, return null
5. Solve this in one pass if possible

Constraints

• The number of nodes in the list is n where 1 ≤ n ≤ 30
• 0 ≤ Node.val ≤ 100
• 1 ≤ k ≤ n (k is always valid)
• Aim for O(n) time complexity and O(1) space complexity

Examples

Example 1:

Input: head = [1,2,3,4,5], k = 2 Output: [1,2,3,5] Explanation: The 2nd node from the end is the node with value 4. After removing it, the list becomes [1,2,3,5].

Example 2:

Input: head = [1], k = 1 Output: [] Explanation: The only node is removed, resulting in an empty list.

Example 3:

Input: head = [1,2], k = 1 Output: [1] Explanation: The last node (with value 2) is removed, leaving only the head node.

Example 4:

Input: head = [1,2], k = 2 Output: [2] Explanation: The head node (with value 1) is removed, and the second node becomes the new head.