What is the time complexity of Double a Number Represented as a Linked List?

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

What data structures are used to solve Double a Number Represented as a Linked List?

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

What are the key insights for solving Double a Number Represented as a Linked List?

Insight 1: Doubling the number is analogous to multiplying it by 2. We need to handle potential carry-overs from right to left, similar to manual multiplication. Insight 2: Since we're dealing with a linked list, we can't easily traverse from right to left. The provided solution cleverly uses the node values themselves to predict and propagate carries from left to right. A node's value greater than 4 means when doubled, the next digit will have a carry-over.

Which companies ask LeetCode 2816. Double a Number Represented as a Linked List in interviews?

This problem is frequently asked at Nvidia.

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Double a Number Represented as a Linked List - LeetCode 2816 Solution

Double a Number Represented as a Linked List - Complete Solution Guide

Double a Number Represented as a Linked List is LeetCode problem 2816, 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 the head of a non-empty linked list representing a non-negative integer without leading zeroes. Return the head of the linked list after doubling it . Example 1: Input: head = [1,8,9] Output: [3,7,8] Explanation: The figure above corresponds to the given linked list which represents the number 189. Hence, the returned linked list represents the number 189 * 2 = 378. Example 2: Input: head = [9,9,9] Output: [1,9,9,8] Explanation: The figure above corresponds to the given linked list

Detailed Explanation

The problem requires you to double a non-negative integer represented as a singly-linked list. Each node in the linked list contains a single digit of the number. The goal is to modify the linked list in-place (if possible or create a new one if necessary) to represent the doubled value of the original number. The input linked list represents a number without leading zeros, except in the case where the number itself is zero. The output should be the head of the modified linked list.

Solution Approach

The solution simulates the doubling process with carry-over propagation. It first checks if the most significant digit (head node's value) is greater than 4. If it is, this means after doubling, we'll have a carry and a new node with value 1 needs to be added at the beginning. The algorithm then iterates through the linked list, doubling each node's value and handling carry-overs. The trick here is to use the value of the next node to know whether to carry over a 1 when doubling the current node's value. If the next node's value is greater than 4, a carry-over will occur from the next node to the current node. The doubled value modulo 10 gives the digit at that node.

Step-by-Step Algorithm

Step 1: Check if the head's value is greater than 4. If it is, create a new node with value 0 and make it the new head, pointing to the original head.
Step 2: Initialize a current pointer to the head of the linked list.
Step 3: Iterate through the linked list using the current pointer. While current is not null:
Step 4: Double the value of the current node.
Step 5: If the current node has a next node and its value is greater than 4, add 1 to the doubled value (carry-over).
Step 6: Update the current node's value to doubled_value % 10.
Step 7: Move the current pointer to the next node.
Step 8: Return the head of the linked list.

Key Insights

Insight 1: Doubling the number is analogous to multiplying it by 2. We need to handle potential carry-overs from right to left, similar to manual multiplication.
Insight 2: Since we're dealing with a linked list, we can't easily traverse from right to left. The provided solution cleverly uses the node values themselves to predict and propagate carries from left to right. A node's value greater than 4 means when doubled, the next digit will have a carry-over.
Insight 3: If the first digit is greater than 4, we'll have a carry-over in the most significant digit, requiring us to create a new head node with value 1 (after the carry-over calculation).

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Linked List, Math, Stack.

Companies

Asked at: Nvidia.

Double a Number Represented as a Linked List | LeetCode 2816

Medium

Linked List Math Stack

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

Solution Code

Available in:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode doubleIt(ListNode head) {
        if (head.val > 4) {
            head = new ListNode(0, head);
        }
        
        ListNode current = head;
        while (current != null) {
            int doubled_value = current.val * 2;
            
            if (current.next != null && current.next.val > 4) {
                doubled_value += 1;
            }
            
            current.val = doubled_value % 10;
            
            current = current.next;
        }
        
        return head;
    }
}

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Problem Info

Description

Problem 2816: Double a Number Represented as a Linked List - Detailed explanation and solution approaches.

Topics

Linked List Math Stack

Companies

Nvidia

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2816

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

8 steps

1
Step 1: Check if the head's value is greater than 4. If it is, create a new node with value 0 and make it the new head, pointing to the original head.
2
Step 2: Initialize a current pointer to the head of the linked list.
3
Step 3: Iterate through the linked list using the current pointer. While current is not null:
4
Step 4: Double the value of the current node.
5
Step 5: If the current node has a next node and its value is greater than 4, add 1 to the doubled value (carry-over).
6
Step 6: Update the current node's value to doubled_value % 10.
7
Step 7: Move the current pointer to the next node.
8
Step 8: Return the head of the linked 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 determined by the single traversal of the linked list. The space complexity is constant because we only use a few extra variables, regardless of the size of the list.

📖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

Convert to Integer, Double, Convert back to Linked List

Convert the linked list to an integer, double the integer, and then convert the doubled integer back to a linked list.

Time: O(n) to convert to integer, O(log(n)) for doubling (can be considered constant), O(m) to convert back to linked list where m is the number of digits in the doubled number. m <= n + 1, so roughly O(n).Space: O(1). Integer variables are used.

Trade-offs: This approach can be simpler to understand initially but may have limitations due to the maximum size of integers that can be stored in a variable. This approach is not optimal due to conversions between LinkedList and Integers.

Stack based reversal

Push all node values onto a stack. Pop and double each value, handling carries. Create a new linked list from the result. Effectively reversing the list temporarily.

Time: O(n) - linear for pushing, popping and creating the list.Space: O(n) - due to the stack.

Trade-offs: Uses additional space for the stack, making the in-place solution preferable if space is a constraint.

✨

Key Insights

1Insight 1: Doubling the number is analogous to multiplying it by 2. We need to handle potential carry-overs from right to left, similar to manual multiplication.
2Insight 2: Since we're dealing with a linked list, we can't easily traverse from right to left. The provided solution cleverly uses the node values themselves to predict and propagate carries from left to right. A node's value greater than 4 means when doubled, the next digit will have a carry-over.
3Insight 3: If the first digit is greater than 4, we'll have a carry-over in the most significant digit, requiring us to create a new head node with value 1 (after the carry-over calculation).

📚

Key Terms Explained

For Students

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Linked List

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

Edge Cases

Edge case 1: Input is [0]. The code correctly doubles it to [0].
Edge case 2: Input is [9,9,9]. The code correctly handles the carry-overs and returns [1,9,9,8].
Edge case 3: Input starts with a value > 4 such as [5, 0, 0]. A new node is created at the beginning to handle the carry.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the carry-over correctly. Forgetting to check for carry and propagate it to the next digit.

✗Common mistake 2: Not considering the edge case where the first digit is greater than 4, leading to an incorrect head node.

✗Common mistake 3: Incorrectly calculating the doubled value or the carry-over. Confusing the modulo operation with integer division.

✅

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: Arbitrary-precision arithmetic calculations where integers can exceed the maximum size supported by standard data types. Linked lists can represent very large numbers.

🏦Real-world scenario 2: Cryptography where large numbers and arithmetic operations on them are fundamental.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

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: September 25, 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

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

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 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 "Double a Number Represented as a 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 math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?