Introduction

In this blog post, we will explore how to solve the LeetCode problem "237" 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

There is a singly-linked list head and we want to delete a node node in it. You are given the node to be deleted node. You will not be given access to the first node of head. All the values of the linked list are unique, and it is guaranteed that the given node node is not the last node in the linked list. Delete the given node. Note that by deleting the node, we do not mean removing it from memory. We mean: The value of the given node should not exist in the linked list. The number of nodes in the linked list should decrease by one. All the values before node should be in the same order. All the values after node should be in the same order. Custom testing: For the input, you should provide the entire linked list head and the node to be given node. node should not be the last node of the list and should be an actual node in the list. We will build the linked list and pass the node to your function. The output will be the entire list after calling your function. Example 1: Input: head = [4,5,1,9], node = 5 Output: [4,1,9] Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function. Example 2: Input: head = [4,5,1,9], node = 1 Output: [4,5,9] Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function. Constraints: The number of the nodes in the given list is in the range [2, 1000]. -1000 <= Node.val <= 1000 The value of each node in the list is unique. The node to be deleted is in the list and is not a tail node.

Explanation

Here's a solution to delete a node in a singly linked list, given access only to the node to be deleted:

• Copy the Next Node's Value: Overwrite the value of the node to be deleted with the value of the node that follows it.

• Update the Next Pointer: Make the next pointer of the node to be deleted point to the node after the next node (essentially skipping over the original next node). This removes the original next node from the list.

• Runtime Complexity: O(1) & Storage Complexity: O(1)

Code

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

def deleteNode(node):
    """
    Deletes a node in a singly linked list given access to only that node.
    """
    node.val = node.next.val
    node.next = node.next.next