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Pigeon hole principle says that if there are N numbers and we are taking N+1 numbers then at least one number bound to get repeated. Similar concept here, let say if the N+M-1 cells are from the same row and same column of the current cell and we take N+M+1 max cells then we are going to find the max and second max excluding all the cells of the current cell's row and column If you find this helpful, please like and subscribe

@@soumyabhattacharjee1051 ok thank you I will rewatch the video

Give a case where its failing

Sure, will keep that in mind

D is kind of a simple breadth-first search, so have not covered it, you can check my solution atcoder.jp/contests/abc361/submissions/55280719 and ask me if there is any doubt

@@soumyabhattacharjee1051 understood the approach, how can we be sure about the tc, can u pls elborate

@@kashishchawla2754 The total number of states are at max O(N2^N) N for choosing the i where i and i+1 are spaces and 2^N for combinations of other. Now for each space the rearrangement is O(N) so overall complexity should not be more than O(N^2*2^N) Please like and subscribe if it is helpful

D is kind of a simple breadth-first search, so have not covered it, you can check my solution atcoder.jp/contests/abc361/submissions/55280719 and ask me if there is any doubt For C you can think about having a subsequence of N-K elements, so we can sort it and consider every subarray of size n-k. Here is my solution atcoder.jp/contests/abc361/submissions/55269760

No (n-1)(n-1) is for choosing anything other than the current j, and 1 when getting (j,j) Please like, share and subscribe if it's helpful

@@soumyabhattacharjee1051 subscribed sir

Try to contruct the string for every value of c and w Here is my python code for ref: atcoder.jp/contests/abc360/submissions/55056440 Please like and subscribe if it is hepful

Thank you so much for the appreciation, it means a lot. Hope you subscribed 😀

Generally cf contests are on weekdays and I do not get time to attend those. If there is some on weekend I can check those. As I am not regular and codeforces contests need grinding, I avoid participating rated. If there is something unrated like div3 in weekends I will definitely try to do it

@@soumyabhattacharjee1051 Understandable.. bro has a job to keep

Thanks for the appreciation, please subscribe if you have not already 😃

Consider the case 1 3 9 15 17 Here we have to ignore 9 which is neither max or min Please subscribe if you find this helpful

Sorry, did not get it. Can you elaborate?

@@soumyabhattacharjee1051 look at the first test case's 7th number input and output

Yes this C broke the flow of the contest, maybe easier C would be better

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Its already there in the channel, please check. Thanks. Please subscribe

Except linkedin

I wanna ask about cp

Hey, lets connect on linkedin

Not necessarily, 2 tricks you can use for the given constraints. Unsigned long long instead of long long to store r=10^s, and then you can use (r^n-1)/(r-1). You can check my code atcoder.jp/contests/abc357/submissions/54462244. Please like and subscribe if it is helpful

@@soumyabhattacharjee1051 got it thanks! Subscribed!

You can check out the article link I have provided in the description for further details, generally if the modulo is prime I use this formula but have not checked the proof yet. Please like and subscribe if you find this helpful

You can check my screencast for this contest ruclips.net/video/-S_n2mNpwqU/видео.html (Please use earphones)

Yes some issue with the voice today, please use earphones for now. Will improve it surely

ans means who is winning. Its initialized to the opposite of the current player. So let say if its 0's turn then ans=1. Now if any one of the possible (i,j) leads us to the 0 then the current player will win. So we are taking bitwise and. Because only one 0 will make the answer 0. Similar for 1, we are taking OR because only one 1 will make the ans 1. If you find this helpful please like, share and subscribe, thanks

@@soumyabhattacharjee1051 oh got it! thanks a lot again! also sir, can you refer some problems of bitmask dp(cf or atcoder) to practice because I just got introduced to it through this video.

ans means who is winning. Its initialized to the opposite of the current player. So let say if its 0's turn then ans=1. Now if any one of the possible (i,j) leads us to the 0 then the current player will win. So we are taking bitwise and. Because only one 0 will make the answer 0. Similar for 1, we are taking OR because only one 1 will make the ans 1. If you find this helpful please like, share and subscribe, thanks