Introduction

In this blog post, we will explore how to solve the LeetCode problem "430" 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 a doubly linked list, which contains nodes that have a next pointer, a previous pointer, and an additional child pointer. This child pointer may or may not point to a separate doubly linked list, also containing these special nodes. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure as shown in the example below. Given the head of the first level of the list, flatten the list so that all the nodes appear in a single-level, doubly linked list. Let curr be a node with a child list. The nodes in the child list should appear after curr and before curr.next in the flattened list. Return the head of the flattened list. The nodes in the list must have all of their child pointers set to null. Example 1: Input: head = [1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12] Output: [1,2,3,7,8,11,12,9,10,4,5,6] Explanation: The multilevel linked list in the input is shown. After flattening the multilevel linked list it becomes: Example 2: Input: head = [1,2,null,3] Output: [1,3,2] Explanation: The multilevel linked list in the input is shown. After flattening the multilevel linked list it becomes: Example 3: Input: head = [] Output: [] Explanation: There could be empty list in the input. Constraints: The number of Nodes will not exceed 1000. 1 <= Node.val <= 105 How the multilevel linked list is represented in test cases: We use the multilevel linked list from Example 1 above: 1---2---3---4---5---6--NULL | 7---8---9---10--NULL | 11--12--NULL The serialization of each level is as follows: [1,2,3,4,5,6,null] [7,8,9,10,null] [11,12,null] To serialize all levels together, we will add nulls in each level to signify no node connects to the upper node of the previous level. The serialization becomes: [1, 2, 3, 4, 5, 6, null] | [null, null, 7, 8, 9, 10, null] | [ null, 11, 12, null] Merging the serialization of each level and removing trailing nulls we obtain: [1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]

Explanation

Here's a breakdown of the solution:

• Depth-First Traversal: Traverse the multilevel linked list using a depth-first approach. When a child node is encountered, flatten the child list before proceeding to the next node in the current level.
• Connect Child and Tail: When flattening a child list, insert the child list after the current node and update the prev pointers accordingly. Find the tail of the flattened child list and connect it to the next node of the current level.

• Nullify Child Pointers: Ensure that all child pointers are set to null after flattening.

• Time Complexity: O(N), where N is the number of nodes in the multilevel linked list.

• Space Complexity: O(1), since the algorithm operates in place and uses constant extra space.

Code

    class Node:
    def __init__(self, val, prev=None, next=None, child=None):
        self.val = val
        self.prev = prev
        self.next = next
        self.child = child


class Solution:
    def flatten(self, head: 'Node') -> 'Node':
        """
        Flattens a multilevel doubly linked list.
        """
        if not head:
            return head

        curr = head

        while curr:
            if curr.child:
                # Flatten the child list
                child_head = curr.child
                child_tail = child_head
                while child_tail.next:
                    child_tail = child_tail.next

                # Connect the child list after curr
                child_head.prev = curr
                child_tail.next = curr.next
                if curr.next:
                    curr.next.prev = child_tail

                curr.next = child_head
                curr.child = None  # Remove child pointer
                curr = child_head  # Continue from the flattened child

            else:
                curr = curr.next

        return head