What is the time complexity of Flatten a Multilevel Doubly Linked List?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Flatten a Multilevel Doubly Linked List?

This problem uses Linked List, Depth-First Search, Doubly-Linked List. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Flatten a Multilevel Doubly Linked List?

Insight 1: The core idea is to traverse the linked list linearly. When a node with a child is encountered, the child list needs to be spliced into the main list. Insight 2: The most efficient way to do this is to find the tail of the child list, which will then be connected to the next node of the parent in the original list.

Which companies ask LeetCode 430. Flatten a Multilevel Doubly Linked List in interviews?

This problem is frequently asked at Arista Networks, SoFi.

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Flatten a Multilevel Doubly Linked List - LeetCode 430 Solution

Flatten a Multilevel Doubly Linked List - Complete Solution Guide

Flatten a Multilevel Doubly Linked List is LeetCode problem 430, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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-leve

Detailed Explanation

The problem asks us to flatten a multilevel doubly linked list into a single-level doubly linked list. The input is the head of the first level of the list. Each node in the list has 'next', 'prev', and 'child' pointers. The 'child' pointer may point to another doubly linked list. The goal is to rearrange the list such that all nodes are in a single level, while preserving the relative order of the nodes. The child list should appear immediately after its parent node and before the parent node's original 'next' node. All child pointers must be set to null after flattening.

Solution Approach

The solution uses an iterative approach to traverse the doubly linked list. When a node with a child is encountered, it finds the tail of the child list. It then updates the 'next' pointer of the current node to point to the head of the child list and updates the 'prev' pointer of the child head. The next pointer of the tail of the child list is updated to point to the next node of the original list, and the prev pointer of next node is updated to the tail node of the child list. Finally, the child pointer of the current node is set to NULL. This process is repeated until the end of the list is reached.

Step-by-Step Algorithm

Step 1: Initialize a 'current_node' to the head of the linked list.
Step 2: Iterate through the linked list using a 'while' loop until 'current_node' becomes null.
Step 3: Inside the loop, check if 'current_node' has a 'child'.
Step 4: If a child exists, store the 'next' node of 'current_node' in 'next_node' before modifying 'current_node.next'.
Step 5: Store the head of the child list in 'child_head'.
Step 6: Find the tail of the child list by traversing the child list to its end.
Step 7: Update the 'next' pointer of 'current_node' to point to 'child_head' and update the 'prev' pointer of 'child_head' to point to 'current_node'.
Step 8: Set the 'child' pointer of 'current_node' to null.
Step 9: Update the 'next' pointer of 'child_tail' to point to 'next_node' and update the 'prev' pointer of 'next_node' to point to 'child_tail'. Handle null case of 'next_node'.
Step 10: Move 'current_node' to its next node in the list.

Key Insights

Insight 1: The core idea is to traverse the linked list linearly. When a node with a child is encountered, the child list needs to be spliced into the main list.
Insight 2: The most efficient way to do this is to find the tail of the child list, which will then be connected to the next node of the parent in the original list.
Insight 3: The manipulation of 'prev' pointers is crucial to maintain the doubly-linked list property. Also, resetting the child pointer to null is mandatory.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Linked List, Depth-First Search, Doubly-Linked List.

Companies

Asked at: Arista Networks, SoFi.

Flatten a Multilevel Doubly Linked List | LeetCode 430

Medium

Linked List Depth-First Search Doubly-Linked List

Time: O(n)Space: O(1)

Solution Code

Available in:

/*
// Definition for a Node.
class Node {
    public int val;
    public Node prev;
    public Node next;
    public Node child;
};
*/

class Solution {
    public Node flatten(Node head) {
        if (head == null) {
            return head;
        }
        
        Node current_node = head;
        while (current_node != null) {
            if (current_node.child != null) {
                Node next_node = current_node.next;
                
                Node child_head = current_node.child;
                
                Node child_tail = child_head;
                while (child_tail.next != null) {
                    child_tail = child_tail.next;
                }
                
                current_node.next = child_head;
                child_head.prev = current_node;
                current_node.child = null;
                
                child_tail.next = next_node;
                if (next_node != null) {
                    next_node.prev = child_tail;
                }
            }
            
            current_node = current_node.next;
            
        }
        return head;
    }
}

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Problem Info

Description

Problem 430: Flatten a Multilevel Doubly Linked List - Detailed explanation and solution approaches.

Topics

Linked List Depth-First Search Doubly-Linked List

Companies

Arista NetworksSoFi

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:430

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Initialize a 'current_node' to the head of the linked list.
2
Step 2: Iterate through the linked list using a 'while' loop until 'current_node' becomes null.
3
Step 3: Inside the loop, check if 'current_node' has a 'child'.
4
Step 4: If a child exists, store the 'next' node of 'current_node' in 'next_node' before modifying 'current_node.next'.
5
Step 5: Store the head of the child list in 'child_head'.
6
Step 6: Find the tail of the child list by traversing the child list to its end.
7
Step 7: Update the 'next' pointer of 'current_node' to point to 'child_head' and update the 'prev' pointer of 'child_head' to point to 'current_node'.
8
Step 8: Set the 'child' pointer of 'current_node' to null.
9
Step 9: Update the 'next' pointer of 'child_tail' to point to 'next_node' and update the 'prev' pointer of 'next_node' to point to 'child_tail'. Handle null case of 'next_node'.
10
Step 10: Move 'current_node' to its next node in the list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because each node is visited and potentially modified a constant number of times. The space complexity is O(1) since the algorithm uses only a few pointers and does not create new nodes or data structures related to n.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Depth-First Search (DFS)

A recursive approach can also be used to flatten the list. It involves recursively flattening the child list and then splicing it into the main list. While conceptually simpler to some, it often has slightly higher overhead due to the recursion stack.

Time: O(n)Space: O(d), where d is the maximum depth of the multilevel list. This is because of the recursion stack.

Trade-offs: Recursive DFS might be more readable but can lead to stack overflow errors for deeply nested lists. The iterative approach is generally preferred for its constant space complexity.

✨

Key Insights

1Insight 1: The core idea is to traverse the linked list linearly. When a node with a child is encountered, the child list needs to be spliced into the main list.
2Insight 2: The most efficient way to do this is to find the tail of the child list, which will then be connected to the next node of the parent in the original list.
3Insight 3: The manipulation of 'prev' pointers is crucial to maintain the doubly-linked list property. Also, resetting the child pointer to null is mandatory.

📚

Key Terms Explained

For Students

Linked List

Elements connected via pointers. Like a chain where each link knows the next.

Edge Cases

Edge case 1: Empty list (head is null). The solution correctly returns null in this case.
Edge case 2: A node has a child, but the next node is null. The solution handles this case by checking if next_node is null before accessing its 'prev' property.
Edge case 3: A child list that is only one node long. The code finds that node as the tail of the child list.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to update the 'prev' pointers when splicing the child list into the main list. This will break the doubly-linked list property.

✗Common mistake 2: Not setting the 'child' pointer of the parent node to null after flattening. The problem explicitly requires this.

✗Common mistake 3: Not handling the case when the 'next' node of the parent node is null after splicing in the child list. This requires a null check before accessing 'next_node.prev'.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Representing hierarchical data structures, such as file systems, organization charts, or nested configurations. Flattening allows for easier traversal and manipulation of the data.

🏦Real-world scenario 2: Game development, where complex object hierarchies are used to manage game entities and their properties. Flattening can simplify rendering and collision detection processes.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

The dummy node technique prevents edge cases. For cycle detection, Floyd's algorithm is the standard. Always handle empty list and single node cases.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Losing reference to nodes when modifying pointers. Use temporary variables to save references.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 10, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

Why use a dummy node in linked list problems?

A dummy node simplifies edge cases where the head might change (insertion/deletion at head) or when building a new list. Initialize dummy.next = head, work with dummy.next throughout, and return dummy.next at the end to handle all cases uniformly.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Flatten a Multilevel Doubly Linked List"?

For this medium-level Linked List problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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