Given the head of a singly linked list and an integer n, remove the nth node from the end of the list (1-indexed) and return the new head. The tricky part isn't finding the node to remove — it's finding the node just before it, since deletion means rewiring a next pointer, and a singly linked list can't look backward on its own. The optimal solution handles this in one pass with the same head startHead Start (n-Gap) TechniqueAdvance one pointer n steps before starting the second one. The n-node gap between them stays constant as both move together, so when the leading pointer reaches the last node, the trailing pointer sits exactly at the predecessor of the nth-from-end node. technique used for finding the kth node from the end: let fast move n steps ahead, then walk both together until fast.next is null — slow lands exactly where it needs to snip. The one edge case is n equaling the list's length, which means the head itself must go.