Leetcode - 19. Remove Nth Node From End of List

You're given a singly linked list. Your task is to remove the nth node from the end and return the updated head of the list.

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💡 Approach

To solve this efficiently in one pass, we use the two-pointer technique:

1. Dummy Node:

   • Add a dummy node pointing to the head. This simplifies edge cases like removing the first node.

2. Two Pointers:

   • Initialize two pointers: left at the dummy node and right at the head.
   • Move right ahead by n steps. This ensures the gap between left and right is n nodes.
   • Now, move both left and right one step at a time until right reaches the end of the list.

3. Remove the Node:

   • At this point, left.next points to the node that needs to be removed.
   • Update left.next = left.next.next to skip the nth node.

4. Return:

   • Return dummy.next which is the new head of the modified list.

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Diagrams

✅ Code (JavaScript)

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */

/**
 * @param {ListNode} head
 * @param {number} n
 * @return {ListNode}
 */
var removeNthFromEnd = function (head, n) {
    let dummy = new ListNode(0, head); // Step 1: Dummy node
    let left = dummy;
    let right = head;

    // Step 2: Move right n steps ahead
    while (n > 0 && right) {
        right = right.next;
        n--;
    }

    // Step 3: Move both pointers until right reaches the end
    while (right) {
        left = left.next;
        right = right.next;
    }

    // Step 4: Remove the nth node
    left.next = left.next.next;

    return dummy.next;
};

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⏱️ Time & Space Complexity

• Time Complexity: O(L)

  • L is the number of nodes in the list.
  • We make a single pass through the list using the two-pointer technique.

• Space Complexity: O(1)

  • No extra space is used apart from a few pointers.

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