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Contest editorials continue kariye please

Bhaiya mei krishna shukla mujhe vote de do mere community post jiska comment link hai uske channel mei uske community post mei it's shive mam se ho banda hai vo mei hu mujhe vote de do pls

Sir, I am in 2nd semester,,when should i apply for an internship?

jump trading opportunity is for 2026 batch or which batch?

thanks sir great

Please bring contest solution codeforces, codechef, leetcode

👍👍👍👍❤️

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Can i say that u find all the subsets in O(n^2)? please correct if i am wrong

Damn what and explanation...🫡

btw its OR, not XOR

how can i get the codeforce id /handle

Thanks for such clear solutions. Please make video solutions for all the upcoming contests.

thankyou very much🙏

bhai tujhe khud toh pta nhi hai , bass answer ratta lga kr aa chuka hai, max draws 9 jab samjane ki baari aayi, toh seedha code par le gaya, aur bataya hi nhi hi 9 draws vaise kaisehue, tune toh 8 hi bataye, aur 9 formula se aaye, oar vaise 9 kaise bane, like kis kis ke beech khelkr 9 bane, ye samjhane ki baat ayi toh tune topic palat diya kuiki tujhe khud nhi samjh aaya , ghochu

explaination for c is crystal clear amazing tutor😊

helpful

problem D what is the reason you instialized east and west as 1

Make a video for 948 (div 2).

As per the Question we need to find the sum of digit differences , so how we got the intution that we need to calculate the freq of each digit in the number and did the calculation afterwards what you have told for the third question smjh ni aya bhaiya please tell me ki ye calculate krke ki how many times a digit is appearing hm ye kaise bta pa rhe , sum of digit differences .. did u got my qUESTION BHAIYA please clear my doubt if it is making sense

You have a legendary observation

Nice solution

bhai kaunsa nasha karke explanation banate ho hume bhi bata do

Have you deleted the leetcode daily videos?

Please continue doing leetcode daily problem videos we are waiting for your videos we love your explaination

First scenario: if(freq[i] != bits[i]) return false; Second Scenario: if(freq[i] > 0 and bits[i] == 0) return false; if(freq[i] == 0 and bits[i] > 0) return false; Both scenarios might work similar but why it is not working for this probelm??

In D problem why we are starting w and e from 1 please give some test case i was not able to find it

Bhai C ki Intuition kaise aayi?

As per the constraints for 4th problem. The solutions should have time complexity O(n). How will O(n(log(n))) work? Can you also do the time complexity analysis for the problem.

why not unordered_map ??

Your explanation is good but coding style is nightmare

how 2*k come?

I did c by finding the lowest value in array, multiolying it by 10 ** 9 and consider it to be the limit., and then calculating the values of every number of array by finding thr ceiil value of arr[i] / lim. Then if sum of new arr b exceeded limit, i returned -1 or else the b array.

That voilet color Advertisment is disturbing us. Please remove it and find some other neat way to put in some other place. Its really disturbing

Super explaination😮😮

thanks for the nice explanation

Nice Explanation sir!

The explanation should be more clear for beginners in codeforces

1:55

Thank you. It's really informative.

I'm unable to understand what exactly this line is going to do in solution of problem B) cout<<(1LL<<bit)<<endl;

this is my brute force solution to the problem D got accepted // !! om namah bhagwate vasudevay !! #include <bits/stdc++.h> using namespace std; int main(){ int t; cin>>t; while(t--){ int n,k;cin>>n>>k; string s;cin>>s; int one=0,zero=0; for(int i=0;i<n;i++){ if(s[i]=='1'){ one++; } else{ zero++; } } if(k==n &&(zero==n || one==n)){ cout<<1<<endl; continue; } int k1=n/k; if(k1%2==0 && zero!=one){ cout<<-1<<endl; continue; } if(k1%2==1 && (zero+k!=one && one+k!=zero)){ cout<<-1<<endl; continue; } if(one%k!=0 || zero%k!=0){ cout<<-1<<endl; continue; } string s1,s2; bool flip = false; int segments = n / k; for (int i = 0; i < segments; ++i) { for (int j = 0; j < k; ++j) { if (flip) { s1.push_back('1'); s2.push_back('0'); } else { s1.push_back('0'); s2.push_back('1'); } } flip = !flip; } //cout int ans=-1; string final=s; reverse(final.begin(), final.end()); if(final==s1 || final== s2){ cout<<n<<endl; continue; } string new1; for(int i=0;i<n-1;i++){ //new1.(s[n-i-1]); new1.insert(0, 1, s[n-i-1]); final.erase(final.begin()); string ans1= new1+ final; if(ans1==s1 ||ans1==s2){ ans=n-i-1; break; } //cout<<ans1<<endl; } cout<<ans<<endl; } }

For 2nd problem you can just do x^ y and return the last set bit in it.

After each contest wait for your videos. Your explanation is much easier and cleaner than any other RUclipsrs. Lots of love and respect for you brother 🥰🥰🥰

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alice bhai hi hai shayad ...