What is the time complexity of Convert Binary Number in a Linked List to Integer?

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

What data structures are used to solve Convert Binary Number in a Linked List to Integer?

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

What are the key insights for solving Convert Binary Number in a Linked List to Integer?

**Iterative MSB-to-LSB Binary Conversion:** The core technique is to build the decimal number by processing binary digits from MSB to LSB. For each new bit encountered, the running total is effectively doubled (shifted left) and then the current bit's value is added. This avoids the need to store all bits or explicitly calculate powers of two. **Efficient Bitwise Operations:** The `(num << 1) | head.val` pattern is the cornerstone. `<< 1` (left shift by one) is a super-fast way to multiply by 2, preparing space for the next bit. `|` (bitwise OR) then neatly inserts the current bit (`0` or `1`) into the newly vacant least significant position. This combination is a standard, highly performant idiom for constructing an integer from a stream of bits.

Which companies ask LeetCode 1290. Convert Binary Number in a Linked List to Integer in interviews?

This problem is frequently asked at MathWorks, Roblox, Workday.

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Convert Binary Number in a Linked List to Integer - LeetCode 1290 Solution

Q: How to solve Convert Binary Number in a Linked List to Integer?

The provided solution leverages a concise and highly efficient iterative approach. It initializes a `num` variable to `0`, which will progressively accumulate the decimal value. The algorithm then traverses the linked list from the head to the tail. For each node's `head.val` (which is either `0` or `1`), it performs a critical operation: `num = (num << 1) | head.val`. Let's break down this operation: `num << 1` performs a bitwise left shift on `num`. This effectively multiplies `num` by `2` and, importantly, shifts all its existing bits one position to the left, making the least significant bit (LSB) position `0`. For example, if `num` was binary `101` (decimal `5`), `num << 1` makes it binary `1010` (decimal `10`). Once this shift creates a `0` at the LSB, `| head.val` performs a bitwise OR operation. Since `head.val` is either `0` or `1`, this operation effectively 'appends' `head.val` into that LSB slot. If `head.val` is `0`, the LSB remains `0`. If `head.val` is `1`, the LSB becomes `1`. This perfectly simulates appending the current binary digit to the right of the growing binary number, continuously updating its decimal equivalent. This process repeats for every node until the list is exhausted, at which point `num` holds the final decimal value.

Q: Which companies ask LeetCode 1290. Convert Binary Number in a Linked List to Integer in interviews?

This problem is frequently asked at MathWorks, Roblox, Workday.

Convert Binary Number in a Linked List to Integer - Complete Solution Guide

Convert Binary Number in a Linked List to Integer is LeetCode problem 1290, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given head which is a reference node to a singly-linked list. The value of each node in the linked list is either 0 or 1 . The linked list holds the binary representation of a number. Return the decimal value of the number in the linked list. The most significant bit is at the head of the linked list. Example 1: Input: head = [1,0,1] Output: 5 Explanation: (101) in base 2 = (5) in base 10 Example 2: Input: head = [0] Output: 0 Constraints: The Linked List is not empty. Number of nodes will not e

Detailed Explanation

You're presented with a singly-linked list where each node's `val` is strictly `0` or `1`. This linked list isn't just holding arbitrary digits; it's representing a binary number. Your objective is to convert this binary representation into its equivalent decimal (base-10) integer value. A crucial piece of information is that the `head` of the linked list contains the *most significant bit* (MSB) of the binary number, meaning the bits are ordered from left to right, just like you'd read a binary number normally.

Solution Approach

The provided solution leverages a concise and highly efficient iterative approach. It initializes a `num` variable to `0`, which will progressively accumulate the decimal value. The algorithm then traverses the linked list from the head to the tail. For each node's `head.val` (which is either `0` or `1`), it performs a critical operation: `num = (num << 1) | head.val`. Let's break down this operation: `num << 1` performs a bitwise left shift on `num`. This effectively multiplies `num` by `2` and, importantly, shifts all its existing bits one position to the left, making the least significant bit (LSB) position `0`. For example, if `num` was binary `101` (decimal `5`), `num << 1` makes it binary `1010` (decimal `10`). Once this shift creates a `0` at the LSB, `| head.val` performs a bitwise OR operation. Since `head.val` is either `0` or `1`, this operation effectively 'appends' `head.val` into that LSB slot. If `head.val` is `0`, the LSB remains `0`. If `head.val` is `1`, the LSB becomes `1`. This perfectly simulates appending the current binary digit to the right of the growing binary number, continuously updating its decimal equivalent. This process repeats for every node until the list is exhausted, at which point `num` holds the final decimal value.

Step-by-Step Algorithm

Step 1: Initialize an integer variable `decimalValue` to 0. This variable will store the accumulating decimal value.
Step 2: Initialize a pointer (e.g., `current`) to the head of the linked list.
Step 3: While the `current` pointer is not null (end of the list):
Step 4: Perform a left bit shift on `decimalValue` by 1 bit: `decimalValue = decimalValue << 1`
Step 5: Perform a bitwise OR operation between `decimalValue` and the current node's value: `decimalValue = decimalValue | current.val`
Step 6: Move the `current` pointer to the next node: `current = current.next`
Step 7: Return the final value of `decimalValue`.

Key Insights

**Iterative MSB-to-LSB Binary Conversion:** The core technique is to build the decimal number by processing binary digits from MSB to LSB. For each new bit encountered, the running total is effectively doubled (shifted left) and then the current bit's value is added. This avoids the need to store all bits or explicitly calculate powers of two.
**Efficient Bitwise Operations:** The `(num << 1) | head.val` pattern is the cornerstone. `<< 1` (left shift by one) is a super-fast way to multiply by 2, preparing space for the next bit. `|` (bitwise OR) then neatly inserts the current bit (`0` or `1`) into the newly vacant least significant position. This combination is a standard, highly performant idiom for constructing an integer from a stream of bits.
**Single Pass, Constant Space:** The solution processes the linked list in a single pass, accumulating the result directly into an integer variable. This achieves optimal time complexity of O(N) where N is the number of nodes, and optimal space complexity of O(1) (excluding the input linked list itself), making it very efficient for potentially large binary numbers.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Linked List, Math.

Companies

Asked at: MathWorks, Roblox, Workday.

Convert Binary Number in a Linked List to Integer | LeetCode 1290

Easy

Linked List Math

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 int getDecimalValue(ListNode head) {
        int decimalValue = 0;
        ListNode current = head;
        while (current != null) {
            decimalValue = (decimalValue << 1) | current.val;
            current = current.next;
        }
        return decimalValue;
    }
}

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Problem Info

Description

Problem 1290: Convert Binary Number in a Linked List to Integer - Detailed explanation and solution approaches.

Topics

Linked List Math

Companies

MathWorksRobloxWorkday

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:1290

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: Initialize an integer variable `decimalValue` to 0. This variable will store the accumulating decimal value.
2
Step 2: Initialize a pointer (e.g., `current`) to the head of the linked list.
3
Step 3: While the `current` pointer is not null (end of the list):
4
Step 4: Perform a left bit shift on `decimalValue` by 1 bit: `decimalValue = decimalValue << 1`
5
Step 5: Perform a bitwise OR operation between `decimalValue` and the current node's value: `decimalValue = decimalValue | current.val`
6
Step 6: Move the `current` pointer to the next node: `current = current.next`
7
Step 7: Return the final value of `decimalValue`.

💡

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 the algorithm iterates through the linked list once, where n is the number of nodes in the list. The space complexity is O(1) because it uses a constant amount of extra space regardless of the input size. Only a few integer variables are used.

📖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

String Conversion

Convert the linked list values to a binary string, then use built-in functions (like `int(binaryString, 2)` in Python) to convert the binary string to an integer.

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

Trade-offs: This approach is less efficient in terms of space complexity because it creates a temporary string of length n. The bit manipulation approach is generally preferred for its efficiency.

✨

Key Insights

1**Iterative MSB-to-LSB Binary Conversion:** The core technique is to build the decimal number by processing binary digits from MSB to LSB. For each new bit encountered, the running total is effectively doubled (shifted left) and then the current bit's value is added. This avoids the need to store all bits or explicitly calculate powers of two.
2**Efficient Bitwise Operations:** The `(num << 1) | head.val` pattern is the cornerstone. `<< 1` (left shift by one) is a super-fast way to multiply by 2, preparing space for the next bit. `|` (bitwise OR) then neatly inserts the current bit (`0` or `1`) into the newly vacant least significant position. This combination is a standard, highly performant idiom for constructing an integer from a stream of bits.
3**Single Pass, Constant Space:** The solution processes the linked list in a single pass, accumulating the result directly into an integer variable. This achieves optimal time complexity of O(N) where N is the number of nodes, and optimal space complexity of O(1) (excluding the input linked list itself), making it very efficient for potentially large binary numbers.

📚

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 linked list. The provided code implicitly handles this as the `while` loop condition will be false immediately, returning 0 which is a correct result for an empty list (though the problem states the list won't be empty).
Edge case 2: Linked list with only one node. The code correctly handles this by performing the shift and OR operation once and returning the node's value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect use of bitwise operators. Beginners might confuse bitwise operators (<<, |) with arithmetic operators. Understanding their behavior is crucial for this problem.

✗Common mistake 2: Not handling edge cases. While the problem statement specifies a non-empty list, robust code should ideally account for potential edge cases such as an empty list for enhanced reliability.

✅

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: Data transmission and storage. Binary linked lists can represent data packets, and efficient conversion to decimal is essential for processing this data.

🏦Real-world scenario 2: Embedded systems and microcontrollers. Resource-constrained environments often use bit manipulation for efficient data representation and processing. This algorithm is directly applicable to such scenarios.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Linked List. It's a great foundation for more complex problems.

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: July 1, 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 14

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

@AlgoMasterJul 7

❓

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.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Convert Binary Number in a Linked List to Integer"?

For this easy-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.

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