19. Remove Nth Node From End of List

At a Glance

Topic: linked-list
Pattern: Fast & Slow Pointers (gap / offset)
Difficulty: Medium
Companies: Amazon, Google, Meta, Apple, Adobe
Frequency: Very High
LeetCode: 19

Problem (one-liner)

Given a singly linked list and nthFromEnd, remove the node that is nthFromEnd steps from the tail and return the head (first node may be removed).

Recognition Cues

“Nth from end” with one pass requirement
Fast pointer leads slow by nthFromEnd gaps
Dummy head simplifies removing the first node

Diagram

At-a-glance flow (replace with problem-specific Mermaid as you refine this note). camelCase node IDs; no spaces in IDs.

Loading diagram…

Approaches

Brute force — two passes: measure length length, remove at index length - nthFromEnd.
Better — one pass with array of pointers — O(n) space.
Optimal — O(n) time / O(1) space — advance fast by nthFromEnd+1 from dummy, then move both until fast hits end; slow.Next is the node to skip.

Optimal Solution

Go

type ListNode struct {
	Val  int
	Next *ListNode
}

func removeNthFromEnd(head *ListNode, nthFromEnd int) *ListNode {
	dummy := &ListNode{Next: head}
	lead := dummy
	for step := 0; step <= nthFromEnd; step++ {
		lead = lead.Next
	}
	trail := dummy
	for lead != nil {
		lead = lead.Next
		trail = trail.Next
	}
	trail.Next = trail.Next.Next
	return dummy.Next
}

JavaScript

class ListNode {
	constructor(val = 0, next = null) {
		this.val = val;
		this.next = next;
	}
}

function removeNthFromEnd(head, nthFromEnd) {
	const dummy = new ListNode(0, head);
	let lead = dummy;
	for (let step = 0; step <= nthFromEnd; step++) {
		lead = lead.next;
	}
	let trail = dummy;
	while (lead !== null) {
		lead = lead.next;
		trail = trail.next;
	}
	trail.next = trail.next.next;
	return dummy.next;
}