Introduction

In this blog post, we will explore how to solve the LeetCode problem "2095" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given the head of a linked list. Delete the middle node, and return the head of the modified linked list. The middle node of a linked list of size n is the ⌊n / 2⌋th node from the start using 0-based indexing, where ⌊x⌋ denotes the largest integer less than or equal to x. For n = 1, 2, 3, 4, and 5, the middle nodes are 0, 1, 1, 2, and 2, respectively. Example 1: Input: head = [1,3,4,7,1,2,6] Output: [1,3,4,1,2,6] Explanation: The above figure represents the given linked list. The indices of the nodes are written below. Since n = 7, node 3 with value 7 is the middle node, which is marked in red. We return the new list after removing this node. Example 2: Input: head = [1,2,3,4] Output: [1,2,4] Explanation: The above figure represents the given linked list. For n = 4, node 2 with value 3 is the middle node, which is marked in red. Example 3: Input: head = [2,1] Output: [2] Explanation: The above figure represents the given linked list. For n = 2, node 1 with value 1 is the middle node, which is marked in red. Node 0 with value 2 is the only node remaining after removing node 1. Constraints: The number of nodes in the list is in the range [1, 105]. 1 <= Node.val <= 105

Explanation

Here's the breakdown of the problem and an efficient solution:

• Approach: Use the fast and slow pointer approach to find the middle node of the linked list. The fast pointer moves two steps at a time, while the slow pointer moves one step at a time. When the fast pointer reaches the end, the slow pointer will be at the middle node. Keep track of the node previous to the slow pointer to facilitate deletion.

• Complexity: O(N) time complexity and O(1) space complexity, where N is the number of nodes in the linked list.

Code

    class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def delete_middle(head: ListNode) -> ListNode:
    """
    Deletes the middle node of a linked list.

    Args:
        head: The head of the linked list.

    Returns:
        The head of the modified linked list.
    """

    if not head or not head.next:
        return None  # Handle empty or single-node list

    slow = head
    fast = head
    prev = None

    while fast and fast.next:
        fast = fast.next.next
        prev = slow
        slow = slow.next

    # Now, slow is pointing to the middle node. Delete it.
    prev.next = slow.next

    return head