Add Binary

Introduction

Hello and welcome! I’m Ming Lee, and today we’re going to cover How to solve LeetCode 67 - Add Binary - in JS/TS ⚡

Same as always we’re gonna solve this, and we’re gonna do it without the big fancy words because this is for regular folx! Keeping it nice and casual round these parts! Movin’ on…

Problem

LeetCode 67 - Add Binary

Given two binary strings a and b, return their sum as a binary string.

Explanation

So this problem does require us to know how binary works, but don’t worry, its fairly simple to get a rudimentary understanding of binary. To start, we need to recognize that it only has 2 numbers, 0 and 1. The numbering system most of us are used to using is called decimal, which has 10 numbers. 0-9. In decimal, when we get to the number 9 and increment, we have to add to a new column, the “tens” column. Once the tens column has used all 10 digits and we need to increment further we introduce a “hundreds” column and so on and so forth. Its actually just as simple when we’re using binary! I’ll produce a table to help explain.

Now that we have a kind of understanding of how binary works we can use that to help us handle this problem. For this we just need to add both of these string’d numbers together. If the number is 0 or 1 that’s totally fine. If the combination adds up to 2, we set that digit to 0 and use a carry over to add to the next digit. If the numbers and carry over add up to 3 then we set the digit to 1 AND send a carry over to the next digit. That might sound a little confusing, but trust me, it’ll make a little more sense when we see it in code so just hang tight for that. Also, because we’re doing addition we typically start our work from right to left. To assist us in making that a little easier, we’ll be reversing both of the strings to make it a little easier to work with. Now lets have a look at this code!

Solution

function addBinary(str1: string, str2: string): string {
    str1 = str1.split('').reverse().join('');
    str2 = str2.split('').reverse().join('');

    let result = '',
        carry = 0;

    for(let i = 0; i < str1.length || i < str2.length; ++i){
        const current = (+str1[i] || 0) + (+str2[i] || 0) + carry;
        result = current % 2 ? 1 + result : 0 + result;
        carry = current > 1 ? 1 : 0;
    }

    return carry ? 1 + result : result;
}

Closing

Nice! So this is definitely one of my most favorite problems. We get to play around with binary, which I super enjoy, and we get to use some fun ternary operators as well! Try giving it a shot and playing around with new ways to handle this problem! You could handle the results in the form of an array or in a string like we’re using here. We could handle things with ternary’s, ifs, or even a switch statement! It’s all programmer’s preference, and there’s nothing wrong with doing things differently than others! You’ve got this! Tomorrow we’ll be workin on LeetCode 69 - Sqrt(x)! See you then!