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Remove Nth Node From End of List

Medium Linked List Front Back Pointer Original on LeetCode

Remove Nth Node From End of List is a medium linked list problem solved with the front back pointer pattern. The best approach, optimal (front and back pointers, one pass), runs in O(L) time and O(1) space. Below are 2 approaches in Java, from two passes up.

Problem

Remove the n-th node from the end of a linked list and return its head.

Follow-up: can you do it in one pass?

Examples

Example 1

Input
head = [1, 2, 3, 4, 5], n = 2
Output
[1, 2, 3, 5]

Example 2

Input
head = [1], n = 1
Output
[]

Constraints

  • 1 <= n <= size <= 30
  • Follow-up: one pass.

Animated walkthrough

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

ApproachTimeSpace
Two passesO(L)O(1)
Optimal (front and back pointers, one pass)O(L)O(1)

1Two passes

TimeO(L)
SpaceO(1)

Count the length L, then remove node L - n + 1 using a dummy node.

  1. Count L.
  2. From dummy, walk L - n steps and unlink the next node.
Java
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        int len = 0;
        for (ListNode p = head; p != null; p = p.next) len++;
        ListNode dummy = new ListNode(0, head), prev = dummy;
        for (int i = 0; i < len - n; i++) prev = prev.next;
        prev.next = prev.next.next;
        return dummy.next;
    }
}

2Optimal (front and back pointers, one pass)

TimeO(L)
SpaceO(1)

Advance front n + 1 steps from a dummy node, then move front and back together until front is null. back is now just before the node to remove.

  1. front = back = dummy; move front n + 1 steps.
  2. While front != null, move both.
  3. back.next = back.next.next.
Java
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0, head), front = dummy, back = dummy;
        for (int i = 0; i <= n; i++) front = front.next;
        while (front != null) { front = front.next; back = back.next; }
        back.next = back.next.next;
        return dummy.next;
    }
}

Edge cases to test

  • Removing the head (n equals the length)
  • Single-node list

Hints

Hint 1

Put the front pointer n steps ahead, then move both. Start from a dummy so the back pointer stops before the node to remove.

FAQ

What is the best time complexity for Remove Nth Node From End of List?

Optimal (front and back pointers, one pass) runs in O(L) time and O(1) extra space.

Which pattern does Remove Nth Node From End of List use?

It is a linked list problem that uses the front back pointer pattern.

Is there a brute force solution for Remove Nth Node From End of List?

Yes. Two passes takes O(L) time and O(1) space. Count the length L, then remove node L - n + 1 using a dummy node.

Which edge cases should I test for Remove Nth Node From End of List?

Removing the head (n equals the length); Single-node list.