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mediumDSA

How do you remove the nth node from the end of a linked list?

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01

Understand the problem

Explain the two-pointer gap technique.

linked-listtwo-pointers
02

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03

Study the solution

Advance a fast pointer n nodes ahead, then move fast and slow together until fast reaches the end — slow now sits just before the target. Use a dummy head so removing the first node is uniform. One pass, O(n).

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04

Read the code

Two-pointer gap
Run Playground
class ListNode:
    def __init__(self, val, nxt=None):
        self.val, self.next = val, nxt

def remove_nth_from_end(head, n):
    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      # unlink target
    return dummy.next


# --- demo ---
def build(vals):
    head = None
    for v in reversed(vals): head = ListNode(v, head)
    return head
def to_list(head):
    out = []
    while head: out.append(head.val); head = head.next
    return out

print(to_list(remove_nth_from_end(build([1, 2, 3, 4, 5]), 2)))   # [1, 2, 3, 5]
05

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