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

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.

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!!!

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) ?

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.

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.

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

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

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

Very Nice Solution, Thanks a lot.

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

Best explanation 😊 Thank you

It was very helpful. Thank you.

very good explanation

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

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

how about Subtraction ?

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(); } }

simply awesome

Thank you

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

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' !...

Nice!

Why did you initialize the sum as carry??

@MrSaurus

Жыл бұрын

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

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

Tnx

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

@alokmehta4631

4 жыл бұрын

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

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

@chemudupatikarthik8

4 ай бұрын

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.

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

3 жыл бұрын

to make it as integer, we do that

@franperez6454

Жыл бұрын

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

Жыл бұрын

@@franperez6454 thank you

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

@weizhou4974

4 жыл бұрын

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

@stayblessed4real

2 жыл бұрын

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

@MrSaurus

Жыл бұрын

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

@unkown657

Жыл бұрын

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.

0:08 system failure

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(); } }

consider editing

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); }