Bro thank you very much.!! Change your mic bro. You have a great potential !.

Code link not work

my Trie solution ``` class TrieNode: def __init__(self): self.child={} self.count=0 class Trie: def __init__(self): self.root=TrieNode() def insert(self,word): dmy=self.root ans=0 for i in word: if i not in dmy.child: dmy.child[i]=TrieNode() else: ans+=dmy.child[i].count dmy=dmy.child[i] dmy.count+=1 return ans n=int(input()) arr=[i for i in input().split()] ans=0 Tree=Trie() for i in arr: ans+=Tree.insert(i) print(ans) ```

Hey Man, Great Explanations. you are bound to go popular soon!. Keep up the good work

What would be the difficulty level of this question as compared to codeforces , any idea what would be its rating .

Great explanation , I like the observation 👍👍

I watched many videos for this problem but it is the short and more intutive, Thanks bro

Great Explanation! Please continue such explanation as main focus should be on intuition and approach rather than code

Excellent explanation.. Keep it up.. Make more editorials for upcoming contests

Your explanations are always crisp and sufficient, thanks a lot

not understand anything at all!

Thanks for this solution dada , during contest I got the idea of using mask but wasn't able to play with the mask correctly.

Saw you competing today after a long time.. Solved A-D and F

Detto code but gives the now , they added so.e extra test cases

Didn't understand the part of the pigeon hole principle. - how did you use it ? Can you explain it in somewhat easy way.

Bhai kya bawasir he ye solution bhi shi nhi he

Great solution!

nice approach of counting the number of segments which will have odd and even number of 1's . there is actually an easier approach which aryanc explained using the prefix xor but it might be a little hard to think about the prefix xor which will make the counting much more simpler

Can you explain solution of problem G?

You explain well, I recomand you to give us all approach from brute force to optimal, If you do it your chanal will grow because no one do this.

nice explanation 👍

very helpful dada ❤

d problem?

thanks

Bro can you explain C and D prb too?

Your explanations are way better than @aryanc's. Thank you for making such good explanatory videos. Keep it up!

thanks for such a simple and great explanation sir!

Thank you sir. I learnt a lot from you.

dp[i][1] = ( 1 + (n - 1) * (n - 1) ) * dp[i - 1] [1] can you explain why you are adding + 1 there. that part is already cover na?

Bro,can you help with B? I am trying to code it in python

underrated channel, love these crisp explanations over those 40 to 60 minute long other video solutions

great solution😊

why did u stop codeforces?

Thank you, finally understood what the problem was trying to say

provide solution for probem G

great explaination

Sir in case one of the values contains an odd number of elements, we have to ignore one of the elements. My intuition was to two create two cases: In the first case we will ignore the smallest element of the set and in the second case we will ignore the largest element of the set, we will take the minimum of the two cases in our answer. However, my approach is failing on some of the test cases.

Bro 23311's output is 19

Could u recommend some dp on bits questions

this contest was more difficult than past contests

Biweekly 4th question

Thanks a lot!

thank u!! I thought the same idea but wasnt able to implement it. Can you also discuss 2 more approaches given in editorial, 1) By SCC and topological sort 2) Something related to depth of directed tree etc.

Great explanation ❤️❤️

Can we do rerooting instead of "in-out dp"

👍

g problem pls, thanks:)

amazing screencast Can you please explain the last question in detail? Thank you

Bhaiya is there any way to connect with you?

Cfbr