🔥🔥✅✅||Simple and Easy solution using python✅✅ || Easy To Understand|| With Explanation
PLEASE UPVOTE IF YOU FIGURED IT OUT\n\n# Intuition\nTo remove the Nth node from the end of a singly linked list, we can utilize two pointers - fast and slow. Th
# PLEASE UPVOTE IF YOU FIGURED IT OUT\n\n# Intuition\nTo remove the Nth node from the end of a singly linked list, we can utilize two pointers - fast and slow. The fast pointer moves ahead by n + 1 steps initially, creating a gap of n nodes between fast and slow. Then, both pointers move together until the fast pointer reaches the end of the list. At this point, the slow pointer will be pointing to the node just before the node to be removed. We can then remove the Nth node by adjusting the next pointer of the node before it.\n\n# Approach\n- Initialize pointers fast and slow to the dummy node.\n- Move the fast pointer n + 1 steps ahead.\n- Move both pointers together until the fast pointer reaches the end.\n- Adjust the next pointer of the node before the Nth node to skip over it.\n- Return the head of the modified list.\n# Complexity\n- Time complexity: O(n), where n is the number of nodes in the linked list.\n- Space complexity: O(1), as we are using only constant extra space for pointers.\n\n# Code\n```python []\nclass Solution:\n    def removeNthFromEnd(self, head, n):\n        if head:\n            # Create a dummy node to handle edge cases\n            dummy = ListNode(0)\n            dummy.next = head\n            fast = dummy\n            slow = dummy\n            # Move fast pointer n+1 steps ahead\n            for _ in range(n + 1):\n                fast = fast.next\n            # Move fast to the end, maintaining the gap of n nodes\n            while fast:\n                fast = fast.next\n                slow = slow.next\n            # Remove the Nth node from the end\n            slow.next = slow.next.next\n            # Return the head of the modified list\n            return dummy.next\n\n```
['Python3']