Q40: Remove Nth Node From End of List. - DSA Patterns (Multi-Lang)

In Simple Words: Advance fast pointer by $N+1$ steps from dummy head. Then move slow and fast together until fast === null. slow.next is now exactly the Nth node from end to delete.

Time: $O(N)$ | Space: $O(1)$

Multi-Language SolutionsPython 3 • JavaScript • TypeScript • Java • C++

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def removeNthFromEnd(head: ListNode | None, n: int) -> ListNode | None:
    dummy = ListNode(0, head)
    slow = fast = dummy
    for _ in range(n + 1):
        fast = fast.next
    while fast:
        slow = slow.next
        fast = fast.next
    slow.next = slow.next.next
    return dummy.next