Codeforces Round 735 Div2 || FaceCam + Commentary + KeyboardCam || A, B, C, D

[@108_adityakumar6]

I had 1 hour 45 minutes and I was totally blank for B.Attempted and got WA on test 2.😶 BTW thank you bhaiya 👍🏻

[@dhayalanm2206]

Brother can u explain the intuition in taking that i, j like that n-200.. Can't understands.. Kindly reply brother

[@DouglasDaMan]

Cant wait to see you reach master, good luck!

[@shadabkouser2446]

Man...u've reached candidate master pretty fast...keep up the good work👍

[@PriyanshAgarwal]

By the way I reached CM in September last year. Hoping to reach Master by this September xD.

[@shadabkouser2446]

@@PriyanshAgarwal even early August 12 2020...😂👍

[@GGxEpsilon]

@@PriyanshAgarwal same 😂

[@movocode]

He has worked hard thats why

[@yashpriyadeepkatta8823]

@20:29, you said that "k times a[i]", why were we able to neglect a[j], can you please explain this point? like why was it not k times a[i] * a[j]?

[@PriyanshAgarwal]

I was looking for the worst case in k * (a[i] | a[j]). So, if a[i] has all bits set in it and it is the maximum number in the array, then taking a[j] and not taking a[j] would mean the same as when you will take an OR you will see that there is no change as the number already has all bits set.

[@yashpriyadeepkatta8823]

@@PriyanshAgarwal Actually i am having hard time understanding these time of concept is it due to lack of practice or i should learn some extra concept?

[@niharjani9611]

Bhaiya can you give a basic c++ code template including all necessary macros , typedef , and fastio statement? If possible

[@PriyanshAgarwal]

Sure, check out my Github repo of CP templates. github.com/Priyansh19077/CP-Templates For particularly the template you can use this: github.com/Priyansh19077/CP-Templates/blob/master/Template.cpp I am pretty sure I am going to get plagiarized someday xD. So many people are using this very same template now.

[@shahriarhabib6452]

bro,which theme do u use in sublime?

[@PriyanshAgarwal]

Brogrammer.

[@prathamchitransh5338]

Coding se jyada typing acchi lg rhi h😅

[@gentleman4615]

How can i talk to you directly and privately?

[@PriyanshAgarwal]

Discord is the place. For 1 on 1 counseling you can take up my commission on memboro memboro.com/priyansh31dec

[@orzAR26]

5:10 i don't know,how many girls would have fallen for this expression!!😂😂😂😂

[@MilindGupta]

Except u none

[@anshshah4927]

B 1500 -1600 today

[@108_adityakumar6]

Yes definitely

[@mloneuskk]

1800 lol

[@vikaspandey8354]

👍

[@ultronxd2860]

Please make a video on time complexity ...... please 🥺

[@PriyanshAgarwal]

I already have a tutorial on time and space complexity on this channel. Please look for a video titled Time Complexity Foobar Beginner class.

[@ultronxd2860]

@@PriyanshAgarwal Thank you i'll check

[@GGxEpsilon]

Problem D was so guessy... Either you get it or you dont

[@PriyanshAgarwal]

Totally agree. I figued it out from the sample test cases. The idea was cool actually.

[@jethalalnhk2409]

At 10:58 you say If I take OR of zero with anything it will be zero. Is it correct

[@PriyanshAgarwal]

I was assuming OR to be AND just after reading the problem 😂. Some time later you will see me realise that in the video.

[@Srlsped]

What is the name of this ide

[@noobie6784]

Stream is nice but please remove the keyboard frontcam.

[@PriyanshAgarwal]

I added it only because a lot of people wanted to see that 😅

[@noobie6784]

@@PriyanshAgarwal Oh okay 😥. Anyways greatwork bro... ❤️️❤️️

[@awwush]

I did B got TLE....

[@PriyanshAgarwal]

Well, you can't submit a normal O(n^2) solution. You can consider the last 100-1000 elements and do an O(n^2) in that many elements only.

[@rishitiwari6200]

@@PriyanshAgarwal how u thought of last 100-1000 I was just thinking of all alternate got wa in pretest 2 and then taken n and n-1 th element but not running on some given tc

[@PriyanshAgarwal]

@@rishitiwari6200 Just look at the first quantity: (i * j) if you pick the last 2 elements then you have n * (n - 1). Now, look at the second quantity k * (a[i] | a[j])... This value can be at max k * n = 100 * n = 100n. So, the minimum answer that you can have from the last 2 elements is n * (n - 1) - 100 n. Now, try to pick the elements n and (n - 2). Now, your best answer can be (n - 2) * n. Clearly, this number will be much smaller than the 1st number for n >= 1000. So, you only have to check for n

[@rishitiwari6200]

@@PriyanshAgarwal thanx for explaining