LeetCode 67. Add Binary Solution Explained - Java

[@curiosity9861]

Very helpful video, but if we dont use carry variable in this program instead after sb.append(sum%2) line we can use sum=sum/2; and after while loop check if sum!=0 if true append it and then returning it will result in 1ms of runtime which is faster than 100% of java code.

[@destodorova8202]

Thank you for taking the time to explain binary with addition examples - very helpful!!

[@iugaialeksei2108]

You are awesome man, so simple and clear explanation for leetcode tasks I've ever seen on youtube. Also, your solutions are always simple and clean.

[@leezhenjian7451]

Hi Nick, thanks for the clear explanation. I was having trouble figuring out how the carry and sum modulus works until i watched ur video.

[@dev-skills]

Below is the solution using Bit Manipulation (same Time Complexity but more space efficient) class Solution { public String addBinary(String a, String b) { int i = a.length() - 1; int j = b.length() - 1; StringBuilder result = new StringBuilder(); boolean carry = false; while(i >= 0 || j >= 0) { boolean sumWithCarry = false; boolean aBit = (i >= 0) ? (a.charAt(i--) - '0' > 0) : false; boolean bBit = (j >= 0) ? (b.charAt(j--) - '0' > 0) : false; boolean sum = aBit ^ bBit; // a ^ b sumWithCarry = sum ^ carry; // a ^ b ^ c carry = (aBit & bBit) | (sum & carry); // a & b | sum & c if (sumWithCarry) result.insert(0, 1); else result.insert(0, 0); } if (carry) { result.insert(0, 1); } return result.toString(); } }

[@darod6098]

Inserting at the end and then reversing is actually better than inserting at the beginning of the string builder with something like sb.insert(0, sum %2) ?

[@ashisenchoudhury162]

You are awesome Nick...Thank you for explaining each and every problem....It is helping me a lot to improve my coding and thinking skills.....Thanks, Love from India!!!

[@RyanMcCoy-p3h]

Honestly the best explanation for this problem on the platform. Thank you :D

[@nikhilpoonia8191]

thank you Nick.....explanation and looks both are on top level

[@rohitrohilla4927]

i wanna tell you i am a beginner and whenever I see your code then There is only a line in my mind 'What a crazy code is this' !...

[@kabilanm9206]

carry ==1 also should be included in while loop condition!!

[@tasneemayham974]

One question: why use a StringBuilder instead of a normal string?

[@akashpal7748]

Best explanation 😊 Thank you

[@satyamgupta6030]

what a nice solution and approach man thanks a ton nick brother. keep making such videos.

[@cengprog9529]

For anyone experiencing an error, here's a better code class Solution { public String addBinary(String a, String b) { StringBuilder sb = new StringBuilder(); int i = a.length() - 1; int j = b.length() - 1; int carry = 0; 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'; sb.append(sum % 2); carry = sum / 2; } return sb.reverse().toString(); } }

[@hande8384]

in case of 11 +11 3 % 2 =0 which will append 0 but 3 in binary is 11 should append 1

[@alokmehta4631]

mod operator gives the remainder and when 3 is divided by 2 the remainder is 1 not 0

[@mashrufehsan]

It was very helpful. Thank you.

[@mohkh7610]

I am a bit confused, let's say the numbers are "11" & "11" Won't the sum add 1 from each first number of these numbers to become 2 then carry is divided by 2, and then sum = carry sum =1 then sum+=1 twice to become 3? I think I perceived something wrongly because I don't think the sum should be 3. or maybe it should be 3 but I don't know how

[@SHSelect]

question: why does '1' - '0' = 1 while '1' + '0' = 97?

[@chemudupatikarthik8]

The reason behind this lies in the way programming languages handle different types of data. When you use the '1' - '0' operation, many programming languages will implicitly convert the characters '1' and '0' into their corresponding ASCII (or Unicode) values before performing the subtraction. In ASCII, the character '1' has a decimal value of 49, and the character '0' has a decimal value of 48. So, '1' - '0' essentially becomes 49 - 48, resulting in 1. However, when you use the '1' + '0' operation, the programming language treats the '1' and '0' as characters to concatenate rather than numbers to add. In ASCII, '1' has a decimal value of 49, and '0' has a decimal value of 48. When you add these characters together, it concatenates them, resulting in the string '10', which in ASCII would represent the character with a decimal value of 97.

[@beibaryslearner]

very good explanation

[@mohankrishna4987]

Very Nice Solution, Thanks a lot.

[@varunrao2931]

Can you do it without using the + operator (only using bit manipulation). FANG has been known to ask this

[@aryanjindal127]

Why did you initialize the sum as carry??

[@MrSaurus]

If I am understanding properly, it is because each time an operation takes place, the carry needs to be sent to sum, for the next operation. So if 1+1 happens, carry would be assigned the value of 2/2, which is 1, and once it loops back, 1 is already assigned to sum because that 1 came from the carry

[@bhushank1234]

Thanks for the solution. You explained it nicely. What is the space and time complexity of this solution?

[@weizhou4974]

The time complexity is o(n) and space complexity is o(1)

[@stayblessed4real]

@@weizhou4974 Can you please explain that why exactly he used string builder?

[@MrSaurus]

@@stayblessed4real I think it's because the method needs to be returned as a string

[@unkown657]

String builder was used for 2 reasons: 1. being able to modify the String, as usual String is immutable object, 2. which also goes back to reason number, which is the reverse function.

[@nhilratha5368]

how about Subtraction ?

[@edgbaston149]

Thank you

[@shauryakapoor2122]

Can anyone explain why we do - '0' in line 11 and line 12? ( sum += a.charAt(i) - '0'); and at ( sum += b.charAt(i) - '0');

[@sushmitagupta6101]

to make it as integer, we do that

[@franperez6454]

in case someone still wonders about this... When using characters in a int type variable (such as sum), they are returned with the ASCII value. a.charAt(1)=49 (ASCII value of 1). So if we want to get the int value of 1 we need to substract the ASCII value of 0 (48). (int) 1-0 = (ASCII) 49-48 = 1 (int) 2-0= (ASCII) 50-48=2 …

[@shauryakapoor2122]

@@franperez6454 thank you

[@feelthestrength9964]

pls solve this write 1 to 5 in binary numbers pls solve it through java coding

[@shritishaw7510]

0:08 system failure

[@akware977]

simply awesome

[@Sobioytccc]

public static String addBinary(String a, String b){ BigInteger firstNum = new BigInteger(a, 2); BigInteger secondNum = new BigInteger(b, 2); BigInteger sum = firstNum.add(secondNum); return sum.toString(2); }

[@raihanoorfarid4973]

Tnx

[@topsiterings]

Nice!

[@ameyamahadevgonal8130]

consider editing