Bit Manipulation

[@LarisaAB]

this is the only video that actually helped me understand bit manipulation, thank you a lot

[@aronpop1447]

Spoiler alert! Exercise question at the end: The idea is to XOR the two numbers and then count how many 1's does the XOR'ed number has. This works because XOR gives 1 only if one bit is 0 and one is 1, so basically XOR solves our problem. We are faced with the task to count how many 1's does a number has. For that you initialize a counter variable to 0 which will hold the solution, then you add to this variable XOR'ed number & 1 then right shift by 1. You do this while your number is greater then 0.

[@HanifCarroll]

Since others have posted solutions to the last exercise, I'll post my alternative solution: def compare_bits(n1, n2): count = 0 while n1 or n2: if n1 & 1 != n2 & 1: count += 1 n1, n2 = n1 >> 1, n2 >> 1 return count

[@griffintoal4410]

im a bit lost :)

[@turn1210]

Nice

[@chhasnainazam7948]

hahahaahha me too :D

[@MAA-op4gw]

Thank you so much.

[@jagannathanrajagopalan2941]

why not use XOR to clear a bit? any reason? i mean 1

[@adibhargava7055]

But what if the starting bit is a 1?

[@ytdl]

Because if you try to clear a bit that is 0 / already cleared it will return a 1.

[@adibhargava7055]

You can also say floor function instead of round down to negative infinity @12:05

[@ytdl]

Never thought i'd see ster teaching bit manipulation

[@cschipg4688]

This guy is so chill presenting. how do you get that relaxed mate

[@maheshj01]

My two cents - Think of it like a normal conversation - and keep your focus on the topic

[@Paradise-kv7fn]

A easier way for the problem of checking a bit would be return (1

[@animodium2670]

This can be simplified like so: return (1

[@LostInNeverland128]

ice town costs ice clown his town crown

[@UdaraAlwis]

yessss! this exactly! I believe only a few will get this reference xD

[@eirikolsnes]

Thank you! Nice to see how giving all students time to respond resulted in valuable input.

[@_sayan_roy_]

24:12 , More intuitive for me is:- int modBit(x,pos,state){ mask1 = 1

[@_sayan_roy_]

+l0ad1 Thanks a lot... :) Now,I do.

[@arunsiddharth5184]

Here you go: int result(int num,int num2){ int x =num ^ num2; int count=0; while(x){ x=(x)&(x-1); count++; } return count; }

[@AviPars]

How would I flip all the bytes ?

[@ahmedramzy892]

great work 🥰

[@satadhi]

0:18 i mean seriously !

[@satadhi]

btw cool videos

[@sergioropo3019]

Very informative and well done. Thank you.

[@iXNomad]

Why do you use -state if you can just write: x = (x & ~(1

[@abdullahnoman8618]

you could use the xor operator in clearing the bit

[@aronpop1447]

Couldnt have you used the ^ operator for clearing the bits? Exact same code except return x ^ mask?

[@aaryandhakal6098]

Might be a little late or u might have even figured ti out but i did use an XOR and the problem is clear_bit(5,1) presents as a counter example which is why XOR is just for flipping as he showed later but seeing as how u solved his last problem u might have watched the entire video and then found that why XOR doesnt work on ure own but still just wanted to help :D

[@olivergreen6436]

Thanks it's very informative

[@panmacabre9895]

that's freaking amazing

[@yamamarques27]

//Write a function to count the numnber of bits that are different between two numbers // Time complexity(1); Space complexity(1) public static void numberOfBitsDifferent(int value1, int value2){ int combination = (value1 ^ value2); int count = 0; for (int i=0; i < 32; i++){ int mask = 1 0)? 1 : 0; } System.out.println("Number of different bits: "+ count); }

[@gauravsoni5974]

What is the mask in terms of bit manipulation?

[@dtm7743]

predefined set of bits used to pick and choose which specific bits will be modified by subsequent operations

[@tushararora347]

What's that sound at 22:41?

[@sergioropo3019]

Obviously is cartoon character exclamation.

[@II_superluminal_II]

now i get brainfuck holy shit, brainfuck is fast as fuck and now I know why.......

[@tauseef10s]

//count number of different bits #include using namespace std; int main(){ // your code goes here int number1, number2, result, count=0; cin >> number1; cin >> number2; result = number1 ^ number2; while (result){ count += result & 1; result = result >> 1; } cout

[@oliviazhai1831]

This is very helpful!!! Thanks

[@mr.anonymous6098]

Great video! Highly recommend it. Wish he was my teacher

[@nishajakhar5867]

You look damn awesome....And even loved your teaching style..

[@vaibhavtiwari6992]

TF LOL

[@cristianrestrepolopez976]

Thank you so much this video was extremely helpful :)

[@pedroalonsoms]

such a good video

[@cadupoles]

Thank you

[@ashishpatel0720]

very good video thank you very much

[@TDLRest]

thanks so much for this! you're awesome teacher

[@vaibhavtiwari6992]

THIS IS GOOD FR SOMEONE WHO ALREADY KNOWS THIS STUFF.

[@gigamilk6981]

This guys pretty cute

[@gigamilk6981]

I take it back hes very cute

[@jonathancao1396]

i love you