LeetCode 19: Remove Nth Node From End of List — Python Solution

Solve LeetCode 19: Remove Nth Node From End of List in Python with a fixed-gap pointers approach. The key is to make the state invariant explicit, so the implementation and complexity follow naturally.

This guide paraphrases the task and does not reproduce LeetCode’s prompt. Use the official page for the complete statement, examples, constraints, and submission runner.

DifficultyMedium
TopicLinked List
Reusable patternfixed-gap pointers
ComplexityO(n) time and O(1) extra space

What the problem is testing

Advance fast n steps from a dummy node, then move fast and slow together until fast reaches the end; slow precedes the target.

Algorithm

  1. Advance fast n steps from a dummy node, then move fast and slow together until fast reaches the end; slow precedes the target.
  2. Maintain this invariant: The pointers remain n nodes apart, so the slow pointer stops immediately before the nth node from the end.
  3. Continue until every input item or reachable state has been resolved, then return the accumulated result.

Python solution

LeetCode provides the list, tree, or graph node definition used by the method.

from collections import Counter, defaultdict, deque, OrderedDict
import random

class Solution:
    def removeNthFromEnd(self, head, n):
        dummy = ListNode(0, head)
        fast = slow = dummy
        for _ in range(n):
            fast = fast.next
        while fast.next:
            fast, slow = fast.next, slow.next
        slow.next = slow.next.next
        return dummy.next

Why this is correct

The proof follows the maintained state: The pointers remain n nodes apart, so the slow pointer stops immediately before the nth node from the end. Each iteration preserves that claim while permanently resolving at least one position, node, interval, or search state. When the loop or recursion ends, every candidate required by the problem has therefore been included or ruled out, so the returned value is correct.

Complexity

O(n) time and O(1) extra space. The stated auxiliary space excludes the returned output unless the output is the data structure being built.

Edge cases

The dummy node handles removing the original head.

Tested reference code

This implementation is included in the site’s downloadable 100-solution Python library. The complete suite compiles every solution and runs a behavioral assertion for every problem before publication.


Browse the searchable 100 LeetCode Python Solutions hub. Previous: 25. Reverse Nodes in k-Group · Next: 82. Remove Duplicates from Sorted List II