Remove Nth Node From End of List

Problem Statement: Given the head of a linked list, remove the nth node from the end of the list and return its head.

Example 1:

Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]

Constraints:

• The number of nodes in the list is sz.
• 1 <= sz <= 30
• 0 <= Node.val <= 100
• 1 <= n <= sz

Source: https://leetcode.com/problems/remove-nth-node-from-end-of-list/

Asked in: Facebook Amazon ,Apple  Netflix Google

Leetcode Difficulty: Medium

Code:

    def removeNthFromEnd(self, head, n):
        """
        head: ListNode
        n: int
        return: ListNode
        """
        fastptr=head
        slowptr=head
        prev=None
        count=1
        
        while count!=n:
            fastptr=fastptr.next
            count+=1
        
        while fastptr.next:
            fastptr=fastptr.next
            prev=slowptr
            slowptr=slowptr.next
            
        
        if prev:
            prev.next=slowptr.next
        else:
            head = slowptr.next
        return head

Thought Process / Explanation:

You can always find the length first and then iterate (L-n) positions from start. But that would require you to do 2 passes on the entire list. So for a single pass, clearly, we need to maintain 2 ptrs (Fast and Slow pointers are very useful in solving LL questions) One starts at the nth position from start and another from Zeroth. By the time ahead one reaches NULL, the previous one will reach the nth position from the end.

Thank You!