Remove Nth Node From End Of List Leetcode Problem 19 [Python Solution]

Welcome to another exciting coding tutorial.

In this post, we'll tackle the "Remove Nth Node from the End of List" LeetCode problem.

We'll provide a step-by-step Python solution and delve into the reasoning behind our approach.

So, let's get started!

Problem Overview

The problem statement is quite straightforward: given the head of a linked list, we need to remove the nth node from the end of the list and return the updated list.

To illustrate, consider the following example:

Example 1:

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

In this case, we want to remove the node with a value of 3, which is the second node from the end of the list.

After removal, the list becomes [1,2,3,5].

Understanding the Constraints

Before we dive into the solution, let's understand the problem constraints:

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

Now that we've grasped the problem statement and constraints, let's explore our efficient Python solution.

Efficient Python Code Solution

We can efficiently solve this problem by using two pointers – a left pointer and a right pointer.

The left pointer will be initially set to a dummy node, and the right pointer will be shifted by n positions to create a gap of n between them.

We'll then advance both pointers until the right pointer reaches the end of the list.

At this point, the left pointer will be at the node we want to remove, and we can update the list accordingly.

Here's the Python code for our efficient solution:

def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
    # Create a dummy node and set it as the next of the head
    dummy = ListNode(0, head)
    left = dummy
    right = head

    # Shift the right pointer by n positions
    while n > 0:
        right = right.next
        n -= 1

    # Advance both pointers until right reaches the end
    while right:
        left = left.next
        right = right.next

    # Delete the node by updating the left pointer's next
    left.next = left.next.next

    # Return the updated list (dummy's next)
    return dummy.next

Time and Space Complexity

• Time Complexity: Our approach runs in O(n) time, where n is the number of nodes in the list.

This is because we traverse the list once using two pointers.
– Space Complexity: We use a constant amount of extra space, resulting in O(1) space complexity.

Reasoning Behind Our Approach

The reasoning behind this approach is quite simple and intuitive.

We aim to find the node to remove by maintaining a gap of n nodes between two pointers.

By doing so, we can efficiently identify the node we need to delete without having to reverse the entire linked list.

The use of a dummy node simplifies the edge cases when the head itself needs to be removed.

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Conclusion

In this blog post, we tackled the "Remove Nth Node from the End of List" LeetCode problem.

We provided an efficient Python solution, discussed the time and space complexity, and explained the reasoning behind our approach.

This solution helps beginners understand a common technique for working with linked lists.

We hope you found this post helpful and insightful.

If you have any questions, suggestions, or want to share your thoughts, please don't hesitate to comment.

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Stay tuned for more exciting coding tutorials and problem-solving strategies.

For the original problem and further discussions, you can visit the LeetCode problem page.

Happy coding!