What is the time complexity of Copy List with Random Pointer?

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

What data structures are used to solve Copy List with Random Pointer?

This problem uses Hash Table, Linked List. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Copy List with Random Pointer?

Insight 1: A simple iterative approach of copying node values will not suffice because the 'random' pointers need to point to the *copied* nodes, not the original ones. Insight 2: A hash table (or dictionary) is crucial to map original nodes to their corresponding copied nodes. This allows you to easily retrieve the copied node when setting the 'next' and 'random' pointers of the copied list.

Which companies ask LeetCode 138. Copy List with Random Pointer in interviews?

This problem is frequently asked at Docusign, Flipkart, Google, Intel, Mobileye and 12 more companies.

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Copy List with Random Pointer - LeetCode 138 Solution

Copy List with Random Pointer - Complete Solution Guide

Copy List with Random Pointer is LeetCode problem 138, 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

A linked list of length n is given such that each node contains an additional random pointer, which could point to any node in the list, or null . Construct a deep copy of the list. The deep copy should consist of exactly n brand new nodes, where each new node has its value set to the value of its corresponding original node. Both the next and random pointer of the new nodes should point to new nodes in the copied list such that the pointers in the original list and copied list represent the sam

Detailed Explanation

The problem requires creating a deep copy of a linked list where each node contains a 'val', a 'next' pointer, and a 'random' pointer. The 'random' pointer can point to any node in the original list or be null. A deep copy means creating entirely new nodes with the same values and linking them in the same way as the original list, without any shared nodes between the original and the copy. The function receives the head of the original linked list as input and must return the head of the new, copied linked list.

Solution Approach

The solution uses a two-pass approach and a hash table (or dictionary) to perform a deep copy of the linked list. The first pass creates all the new nodes and stores the mapping between the original nodes and the new nodes in the hash table. The second pass iterates through the original linked list again and uses the hash table to set the 'next' and 'random' pointers of the new nodes, effectively recreating the structure of the original list in the copied list.

Step-by-Step Algorithm

Step 1: Handle the base case: If the input `head` is null, return null.
Step 2: Create a hash table (dictionary) called `old_to_new` to store the mapping between original nodes and their copies. Initialize it with {None: None} or {null: null} to handle null random pointers.
Step 3: Iterate through the original linked list. For each node, create a new node with the same value and store the mapping between the original node and the new node in the `old_to_new` hash table.
Step 4: Iterate through the original linked list again. For each node, retrieve its corresponding copied node from the `old_to_new` hash table.
Step 5: Set the `next` pointer of the copied node to the copied version of the original node's `next` pointer (using the `old_to_new` hash table).
Step 6: Set the `random` pointer of the copied node to the copied version of the original node's `random` pointer (using the `old_to_new` hash table).
Step 7: Return the copied version of the original head node from the `old_to_new` hash table.

Key Insights

Insight 1: A simple iterative approach of copying node values will not suffice because the 'random' pointers need to point to the *copied* nodes, not the original ones.
Insight 2: A hash table (or dictionary) is crucial to map original nodes to their corresponding copied nodes. This allows you to easily retrieve the copied node when setting the 'next' and 'random' pointers of the copied list.
Insight 3: Handling the null 'random' pointers correctly is important. They should be mapped to null in the copied list.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, Linked List.

Companies

Asked at: Docusign, Flipkart, Google, Intel, Mobileye, Morgan Stanley, Nvidia, Oracle, PhonePe, Snowflake, Uber, VMware, Walmart Labs, Wix, Yahoo, eBay, oyo.

Copy List with Random Pointer | LeetCode 138

Medium

Hash Table Linked List

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

Solution Code

Available in:

/*
// Definition for a Node.
class Node {
    int val;
    Node next;
    Node random;

    public Node(int val) {
        this.val = val;
        this.next = null;
        this.random = null;
    }
}
*/

import java.util.HashMap;

class Solution {
    public Node copyRandomList(Node head) {
        if (head == null) {
            return null;
        }
        
        HashMap<Node, Node> oldToNew = new HashMap<>();
        oldToNew.put(null, null);
        
        Node current = head;
        while (current != null) {
            Node copy = new Node(current.val);
            oldToNew.put(current, copy);
            current = current.next;
        }
        
        current = head;
        while (current != null) {
            Node copy = oldToNew.get(current);
            copy.next = oldToNew.get(current.next);
            copy.random = oldToNew.get(current.random);
            current = current.next;
        }
        
        return oldToNew.get(head);
    }
}

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Problem Info

Description

Problem 138: Copy List with Random Pointer - Detailed explanation and solution approaches.

Topics

Hash Table Linked List

Companies

DocusignFlipkartGoogleIntelMobileyeMorgan StanleyNvidiaOracle

Stats

Difficulty:Medium

Languages:4

Companies:17

Problem #:138

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

7 steps

1
Step 1: Handle the base case: If the input `head` is null, return null.
2
Step 2: Create a hash table (dictionary) called `old_to_new` to store the mapping between original nodes and their copies. Initialize it with {None: None} or {null: null} to handle null random pointers.
3
Step 3: Iterate through the original linked list. For each node, create a new node with the same value and store the mapping between the original node and the new node in the `old_to_new` hash table.
4
Step 4: Iterate through the original linked list again. For each node, retrieve its corresponding copied node from the `old_to_new` hash table.
5
Step 5: Set the `next` pointer of the copied node to the copied version of the original node's `next` pointer (using the `old_to_new` hash table).
6
Step 6: Set the `random` pointer of the copied node to the copied version of the original node's `random` pointer (using the `old_to_new` hash table).
7
Step 7: Return the copied version of the original head node from the `old_to_new` hash table.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The algorithm involves two linear traversals of the linked list. The hash map `old_to_new` stores a mapping for each node in the list. Therefore, the time and space complexities are both directly proportional to the number of nodes.

📖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

Interweaving Nodes

Create a copy of each node and insert it immediately after the original node. Then, iterate through the interwoven list, setting the random pointers of the copied nodes. Finally, separate the original and copied lists.

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

Trade-offs: This approach has constant space complexity but modifies the original linked list in place. If modifying the original list is not allowed, it cannot be used. It's also slightly more complex to implement than the hash table approach.

✨

Key Insights

1Insight 1: A simple iterative approach of copying node values will not suffice because the 'random' pointers need to point to the *copied* nodes, not the original ones.
2Insight 2: A hash table (or dictionary) is crucial to map original nodes to their corresponding copied nodes. This allows you to easily retrieve the copied node when setting the 'next' and 'random' pointers of the copied list.
3Insight 3: Handling the null 'random' pointers correctly is important. They should be mapped to null in the copied list.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Linked List

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

Edge Cases

Edge case 1: Empty linked list (head is null). The solution handles this by returning null immediately, avoiding any further processing.
Edge case 2: Nodes with null 'random' pointers. The hash table is initialized with a mapping for 'None' or 'null', ensuring these pointers are correctly handled in the copied list.
Edge case 3: Self-referential random pointers (a node's random pointer points to itself). The hash table correctly maps the original node to its copy, so the copied node's random pointer will point to its copy correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not creating completely new nodes and instead pointing to the original nodes, resulting in a shallow copy rather than a deep copy.

✗Common mistake 2: Forgetting to handle the case where the `random` pointer is null, which can lead to errors when trying to access the copied version of a null pointer.

✗Common mistake 3: Not correctly mapping the head node to its copy, leading to returning the wrong head of the copied linked list.

✅

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: Serialization and deserialization of complex data structures. When serializing an object graph with circular references, a deep copy is needed to ensure that the serialized data contains all the necessary information without infinite loops.

🏦Real-world scenario 2: Implementing undo/redo functionality in software. A deep copy of the application state is often needed to preserve the previous state before applying a new operation, allowing the user to revert to the previous state if needed.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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: October 17, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 "Copy List with Random Pointer"?

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

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