What is the time complexity of Add Binary?

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

What data structures are used to solve Add Binary?

This problem uses Math, String, Bit Manipulation, Simulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Add Binary?

Simulate manual binary addition: Iterate from right-to-left (least significant bit to most significant bit), just like paper-and-pencil arithmetic, calculating each sum and managing a carry-over. Explicitly handle carry propagation: The core logic for each bit is `(digit_a + digit_b + carry_in) % 2` for the current result bit, and `(digit_a + digit_b + carry_in) // 2` for the `carry_out` to the next position.

Which companies ask LeetCode 67. Add Binary in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Google and 6 more companies.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Add Binary - LeetCode 67 Solution

Q: How to solve Add Binary?

The solution elegantly mirrors how we'd manually add binary numbers with pencil and paper. We iterate from the least significant bit (rightmost character) to the most significant bit (leftmost character), keeping track of a `carry`. For each position, we sum the bits from `a`, the bits from `b`, and the `carry` from the previous position. The current bit of our result is the sum modulo 2 (e.g., 0+0=0, 0+1=1, 1+1=0 with a carry, 1+1+1=1 with a carry). The new `carry` for the next position is the sum integer-divided by 2. This process continues until all bits are processed, and finally, any leftover `carry` is prepended to the result. Padding the shorter string with leading zeros (`zfill`) simplifies the loop by ensuring both strings have the same effective length, preventing out-of-bounds access and making the alignment straightforward.

Add Binary - Complete Solution Guide

Add Binary is LeetCode problem 67, 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 two binary strings a and b , return their sum as a binary string . Example 1: Input: a = "11", b = "1" Output: "100" Example 2: Input: a = "1010", b = "1011" Output: "10101" Constraints: 1 <= a.length, b.length <= 10 4 a and b consist only of '0' or '1' characters. Each string does not contain leading zeros except for the zero itself.

Detailed Explanation

At its heart, this problem asks us to perform binary addition, just like we learned decimal addition in school, but with only two digits: '0' and '1'. The catch? Our binary numbers are provided as strings, and they can be quite long – up to 10,000 characters! This immediately tells us we can't simply convert them to integers, add them, and convert back, as standard integer types (like Python's `int` or Java's `long`) would overflow for such massive numbers.

Solution Approach

The solution elegantly mirrors how we'd manually add binary numbers with pencil and paper. We iterate from the least significant bit (rightmost character) to the most significant bit (leftmost character), keeping track of a `carry`. For each position, we sum the bits from `a`, the bits from `b`, and the `carry` from the previous position. The current bit of our result is the sum modulo 2 (e.g., 0+0=0, 0+1=1, 1+1=0 with a carry, 1+1+1=1 with a carry). The new `carry` for the next position is the sum integer-divided by 2. This process continues until all bits are processed, and finally, any leftover `carry` is prepended to the result. Padding the shorter string with leading zeros (`zfill`) simplifies the loop by ensuring both strings have the same effective length, preventing out-of-bounds access and making the alignment straightforward.

Step-by-Step Algorithm

Step 1: Initialize a carry variable to 0 and an empty string to store the result.
Step 2: Iterate through the binary strings from right to left (least significant bit to most significant bit).
Step 3: For each bit position, add the bits from both input strings (converted to integers) and the carry.
Step 4: The remainder when the sum is divided by 2 represents the current bit of the result. Add this bit to the result string.
Step 5: The quotient when the sum is divided by 2 is the new carry.
Step 6: After the loop, if the carry is 1, add a leading '1' to the result.
Step 7: Return the reversed result string (because we built the string from right to left).

Key Insights

Simulate manual binary addition: Iterate from right-to-left (least significant bit to most significant bit), just like paper-and-pencil arithmetic, calculating each sum and managing a carry-over.
Explicitly handle carry propagation: The core logic for each bit is `(digit_a + digit_b + carry_in) % 2` for the current result bit, and `(digit_a + digit_b + carry_in) // 2` for the `carry_out` to the next position.
Don't forget the final carry: After processing all corresponding digits, a lingering `carry` (if it's '1') must be prepended to the entire result, as seen in `"11" + "1" = "100"` where the '1' is the final carry.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Math, String, Bit Manipulation, Simulation.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Google, Meta, Microsoft, SIG, Uber, Walmart Labs, tcs.

Add Binary | LeetCode 67

Easy

Math String Bit Manipulation Simulation

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

Solution Code

Available in:

class Solution {
    public String addBinary(String a, String b) {
        StringBuilder result = new StringBuilder();
        int carry = 0;
        int i = a.length() - 1;
        int j = b.length() - 1;

        while (i >= 0 || j >= 0 || carry > 0) {
            int sum = carry;
            if (i >= 0) {
                sum += a.charAt(i--) - '0';
            }
            if (j >= 0) {
                sum += b.charAt(j--) - '0';
            }

            result.append(sum % 2);
            carry = sum / 2;
        }

        return result.reverse().toString();
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 67: Add Binary - Detailed explanation and solution approaches.

Topics

Math String Bit Manipulation Simulation

Companies

AdobeAmazonAppleBloombergGoogleMetaMicrosoftSIG

Stats

Difficulty:Easy

Languages:4

Companies:11

Problem #:67

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 a carry variable to 0 and an empty string to store the result.
2
Step 2: Iterate through the binary strings from right to left (least significant bit to most significant bit).
3
Step 3: For each bit position, add the bits from both input strings (converted to integers) and the carry.
4
Step 4: The remainder when the sum is divided by 2 represents the current bit of the result. Add this bit to the result string.
5
Step 5: The quotient when the sum is divided by 2 is the new carry.
6
Step 6: After the loop, if the carry is 1, add a leading '1' to the result.
7
Step 7: Return the reversed result string (because we built the string from right to left).

💡

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 iterates through the strings once. The length of the result string is at most n+1 (due to a possible carry). All other operations (addition, modulo, etc.) take constant time.

📖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

Bit Manipulation using Integer Addition

Convert the binary strings to integers using built-in functions (like `int(a, 2)` in Python), add them, and then convert the result back to a binary string using a built-in function.

Time: O(n), where n is the maximum length of the input strings (although converting a number to binary might seem O(logn), this function is highly optimized in most languages and tends to have better performance)Space: O(1), excluding the space used for the input and output.

Trade-offs: This approach might be simpler and potentially slightly faster due to optimized integer arithmetic functions provided by programming languages, especially for longer inputs. However, it might rely on implicit input size limits (integers in programming languages have maximum values).

✨

Key Insights

1Simulate manual binary addition: Iterate from right-to-left (least significant bit to most significant bit), just like paper-and-pencil arithmetic, calculating each sum and managing a carry-over.
2Explicitly handle carry propagation: The core logic for each bit is `(digit_a + digit_b + carry_in) % 2` for the current result bit, and `(digit_a + digit_b + carry_in) // 2` for the `carry_out` to the next position.
3Don't forget the final carry: After processing all corresponding digits, a lingering `carry` (if it's '1') must be prepended to the entire result, as seen in `"11" + "1" = "100"` where the '1' is the final carry.

📚

Key Terms Explained

For Students

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: One string is much longer than the other. The solution handles this by padding the shorter string with leading zeros using `zfill` (Python) or implicit zero handling in the while loop (other languages).
Edge case 2: Both strings are "0". The solution correctly returns "0".

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the carry-over correctly. Incorrectly adding bits without considering the carry can lead to incorrect results.

✗Common mistake 2: Inefficient string manipulation. Repeated string concatenation in Python or Java can be slow. Using StringBuilder (Java) or other efficient string manipulation techniques is recommended.

✅

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: Low-level programming and embedded systems where binary arithmetic is directly used to manipulate hardware registers or data.

🏦Real-world scenario 2: Network protocols or communication systems that use binary data representation and require binary arithmetic for data processing or error correction.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary 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: August 7, 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

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

@InterviewProJul 14

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

@CodeNinjaJul 7

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

@TechLeadJun 30

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Add Binary"?

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

🤖 Need help with this problem?