How do you solve Remove Nth Node From End of List on LeetCode?

Remove Nth Node From End of List (LeetCode #19) can be solved using Brute Force or Optimal Solution (Two Pointers). The optimal approach is Optimal Solution (Two Pointers) with O(n) time complexity and O(1) space complexity. Key insight: Using two pointers allows for an efficient solution that only requires a single pass through the linked list.

What is the brute force solution for Remove Nth Node From End of List?

The brute force approach for Remove Nth Node From End of List is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calculating the length of the linked list first, and then identifying the node to be removed based on its position from the start. This is straightforward but inefficient because it requires two passes through the list. The optimal Optimal Solution (Two Pointers) approach improves this to O(n).

What algorithm or data structure is used to solve Remove Nth Node From End of List?

Remove Nth Node From End of List uses the following concepts: Linked List, Two Pointers. The recommended patterns to study are: Hash Map, Array, Two Pointers.

What are the interview tips for Remove Nth Node From End of List?

When you first see this problem, clarify the constraints and edge cases, such as what happens when n equals the length of the list. Communicate your thought process clearly, explaining why you would use two pointers and how it optimizes the solution. Mention edge cases like a single-node list or removing the head node to demonstrate your understanding of the problem.

What are common mistakes when solving Remove Nth Node From End of List?

Students often forget to handle the case where the head needs to be removed, leading to null pointer exceptions. Another common mistake is not properly updating the pointers when removing the node, which can lead to memory leaks or incorrect list structures.

#19

Remove Nth Node From End of List

Medium

Linked ListTwo PointersHash MapArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution (Two Pointers)★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Using two pointers allows us to find the nth node from the end in a single pass. By advancing one pointer n steps ahead, we can then move both pointers until the first pointer reaches the end, at which point the second pointer will be at the node just before the one we want to remove.

⚙️

Algorithm

4 steps

1Step 1: Initialize two pointers, both starting at the head.
2Step 2: Move the first pointer n steps ahead.
3Step 3: Move both pointers until the first pointer reaches the end of the list.
4Step 4: The second pointer will now be at the node just before the one to be removed. Adjust the pointers to remove the target node.

solution.py26 lines

1# Full working Python code with comments
2class ListNode:
3    def __init__(self, val=0, next=None):
4        self.val = val
5        self.next = next
6
7class Solution:
8    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
9        dummy = ListNode(0)
10        dummy.next = head
11        first = dummy
12        second = dummy
13
14        # Step 2: Move first pointer n steps ahead
15        for _ in range(n + 1):
16            first = first.next
17
18        # Step 3: Move both pointers until first reaches the end
19        while first:
20            first = first.next
21            second = second.next
22
23        # Step 4: Remove the nth node
24        second.next = second.next.next
25
26        return dummy.next

ℹ

Complexity note: The time complexity is O(n) because we traverse the list only once. The space complexity is O(1) since we are using a constant amount of extra space.

1Using two pointers allows for an efficient solution that only requires a single pass through the linked list.
2The dummy node pattern helps handle edge cases like removing the head of the list without additional checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.