UTF - 8 Encoding Explained using Bit Manipulation | Bit Manipulation Interview Questions

[@sourin.majumdar]

itni patience k sath samjhaya, Salute hai!! Crystal clear samajh aaya Ek aur testcase batana chahunga. If given array is [255], tab uski binary representation ho jaayegi 11111111. But uska check toh lagaya hi nhi, sidhe false return karna hai. So 3rd else if k baad ek 'else return false' lag jaayega bass kaam khatam

[@prajyotgurav]

I couldn't understand today's Leetcode #393 question. Thanks to your video. I was able to understand and solve the question. 🙌🙌🙌

[@adityarao8803]

Awesome Explanation Sir. I was pretty confused looking and referring to other videos your video was very precise and easy to understand.

[@ritikmishra5210]

Those facing issue in leetcode submission, just add this condition in first if if(rbytes==0) { if ... else if ... . . *else return false* }

[@Pepcoding]

Visit nados.pepcoding.com and sign up to NADOS, for better experience and precisely arranged content. Don't forget to follow us on Instagram instagram.com/pepcoding/ Happy Coding!

[@Shivanai_h]

Thank u ritik😀

[@atulkumarverma2042]

I was here to comment the same 😇

[@akhilesh59]

Very Nice Explanation. Thanks a lot Sir

[@creativegiant148]

Sir kuch case reh gaye hain solution me jaise ki agr pehla number hii 10 se start hoga toh program handle nahi kr paye ga usse Btw rest is great!!

[@mickyman753]

i found the checking logic for remaining bytes in loop beautiful ,sumeet sir great as always

[@shubhamrawat7895]

Wao.. Great logic! I was thinking to use if else for every checking.. Using remainingbytes variable is nice 🔥😮

[@Pepcoding]

Je baat. Mza-mev-jayte. Mze ke hmesha vijay hoti hai. jo fun aa rha ho to ek review daal do yar g.page/Pepcoding/review?rc

[@rahulbhatia3075]

Next level explaination 🙏🙏

[@Pepcoding]

Keep watching!! Share as much as you can

[@rahulbhatia3075]

Sir pls make video on segment a nd Fenwick trees pls

[@pariveshmehta2298]

sir isme aapne ek case consider nahi kiya jisme koi bhi condition satisfy na ho. for example [255] but thanks for the awsome eplaination.🤗

[@prernnajain1532]

In that case we need to return false na

[@balaveeraraghavareddy9764]

Thanks a lot very nice explanation

[@pjagannadham2540]

good explanation sir

[@pareshshikhaliya4702]

best explanation

[@shwetanksingh5438]

Thank You!

[@ankitmindewar3710]

sir, video portal par play nhi ho rhi hai.

[@Pepcoding]

beta will look into that

[@umanggupta9199]

Sir I think the code doesn't handle one case in which one element starts with 110 and next two or more elements start with 10. For that you need to add an else return false in first if case.

[@glorysingh]

Yes I noticed the same bug.

[@aashishjha3223]

Sir, it's not submitting in leetcode. If we put else if(val >> 7) != 0 return false in last in else if then it's submitting. testcase - [255]

[@Elon-musk-007]

it is because if there is anything else apart from these 4 conditions than it will always return false

[@LegitGamer2345]

@@Elon-musk-007 will always return true

[@AmanRaj-nn6xx]

modify the code by adding false condition like this(If you still have the doubt and you survived the pandemic):- if(rembit == 0){ if( (val >> 7) == 0 ){ rembit = 0; } else if( (val >> 5) == 6){ rembit = 1; } else if( (val >> 4) == 14){ rembit = 2; } else if( (val >> 3) == 30){ rembit = 3; } else{ return false; }

[@satyajeetrajupali3147]

Correct Solution: int rbytes = 0; for(int num: data) { if(rbytes == 0){ if((num >> 3) == 0b11110) rbytes = 3; else if ((num >> 4) == 0b1110) rbytes = 2; else if ((num >> 5) == 0b110) rbytes = 1; else if ((num >> 7) == 0b0) rbytes = 0; else return false; } else { if ((num >> 6) != 0b10){ return false; } rbytes--; } } return rbytes == 0;

[@amandixit3555]

sir aap agar formal shirt mei vedio bano ge to jyada viwes aayenge , give it a try !!!

[@Pepcoding]

Yeah, bhot din ho gae. Kal se try karta hun. Shayad nha ke bnaun aur shave regular karoon to bhi fayda hoga. I agree, a teacher must look presentable.

[@code7434]

Sir binary search ke partitions of array ke que ki variations krwado please :)

[@Pepcoding]

it's there in array section

[@code7434]

@@Pepcoding can u share the link , I m asking for problems like painter partition using binary search and its similar variations :).

[@NishantKumar-hu7dq]

We need to handel one more case when rbytes==0 In else part for these type of input 197 130 8 130

[@sanjaymajhi7861]

sir, as per what u said in first video, that if there are 8 bits , then 7 bits represent the number and 1 bit for sign, then why this thing is not there in this question. I am confused at this point. Thank you

[@shivanishree276]

sir can you please explain in English

[@orien1116]

sir what if [255] ?this shouldn't be the else condition when remainingByes == 0!?

[@sailendrachettri8521]

add else int the else if statement class Solution { public: bool validUtf8(vector& data) { int remBytes = 0; for(auto val : data) { if(remBytes == 0) { if( (val >> 7) == 0b0 ) remBytes = 0; else if( (val >> 5) == 0b110) remBytes = 1; else if( (val >> 4) == 0b1110) remBytes = 2; else if( (val >> 3) == 0b11110) remBytes = 3; else return false; } else { if( (val >> 6) == 0b10) remBytes--; else return false; } } if(remBytes == 0) return true; return false; } };

[@JitendraKumar-ti6yd]

Leetcode solutions: public boolean validUtf8(int[] data) { int remBytes = 0; for(int var : data) { if(remBytes == 0) { if((var >> 7) == 0b0) { // 1byte remBytes = 0; } else if((var >> 5) == 0b110) { // 2 byte remBytes = 1; } else if ((var >> 4) == 0b1110) { // 3 byte remBytes = 2; } else if((var >> 3) == 0b11110) { // 4byete remBytes = 3; } else { return false; } } else { if((var >> 6) == 0b10) { remBytes--; } else { return false; } } } if(remBytes == 0) { return true; } else { return false; } }

[@sujitwalmiki3603]

Underrated content🥲