วีดีโอ

How and where do i register and attempt the test

You made such hard concept easy to understand really appriciate your effort

Hi can you please help with the slides

I didn't registered for the Camp, and i don't want to miss any event in future.. is there any discord server or any equivalent group in which i can join ?

What's this? I didn't understand anything 🥲

Where to register

Where to register?

nice

how do we do it iteratively

Great explanation. Kudos to you.

Hi can you share this input which you have used in video for explanation?

At 24:19, i think ans(NR) should be subtracted by cntsz_old(NR) instead of cntsz_new(NR) right?

is there free pdf available for cp handbook?

hi nice video

it was a nice problem

great q

accent sudhar le bsdk

very clear and great explanation. Thanks !!

Hi, could help me find out why my thought is wrong? My idea is for every index i, calculate the max sum of subarrays that are end at index i, so if the max element in the subarray is less than the max sum, the answer is no, otherwise yes. The code is like this: void solve() { int n; cin >> n; vector<ll> a(n); for(int i=0; i<n; i++) cin >> a[i]; ll cur = -1e18, nax = -1e18; for(int i=0; i<n; i++) { if(a[i] >= cur + a[i]) { cur = a[i]; nax = a[i]; } else { cur += a[i]; nax = max(nax, a[i]); } if(nax < cur) { cout << "No" << endl; return; } } cout << "Yes" << endl; }

Man, he literally copied the whole code from the tutorials section. I didn't like this thing.

i guess with prefix array we can also make a greater next element sort of thing to check the conditions like x>0 for then range (i+1,next_greater for ith index)

I can bet on explaining this problem in this comment better than this student. Foremost, one edge case, if n == 1, output -1 and return to the main function because you won't be able to shuffle anything. Simple, right... Then, I declare a map of int and vector<int>, to store the numbers and their respective indices on which they occur in the vector. In easy words, the number entered by the user is the key, and the indices on which the number occurs get stored in the vector (i.e. the value for that particular key). Now, if a specific shoe size occurs a single time, straight away print -1 and return to the main function. Because there would exist one such person, who would get the smaller shoe size(i.e. the unique size), and this would violate the operation's rules. We can only shuffle if each shoe size occurs at least 2 times in the whole input vector. This means, that we can only shuffle the same shoe sizes. Rest is just implementation. Below is the code, which will make all this thing crystal clear to the reader. #include<bits/stdc++.h> #define ll long long int #define cy cout<<"YES " ; #define cn cout<<"NO " ; #define mod 1000000007 using namespace std ; void solve(){ ll n ; cin>>n ; map<ll , vector<ll>> mp ; vector<ll> z ; for(ll i = 0 ; i < n ; i++){ ll x ; cin>>x ; mp[x].push_back(i) ; } if(n == 1){ cout<<"-1 " ; return ; } vector<ll> v ; for(auto it: mp){ if(it.second.size() == 1){ cout<<"-1 " ; return ; } ll j = it.second[0] ; for(ll i = 1 ; i < it.second.size() ; i++) v.push_back(it.second[i] + 1) ; v.push_back(j + 1) ; } for(auto it: v) cout<<it<<" "; cout<<endl ; } int main(){ ll t ; cin>>t ; while(t--) solve() ; } //Thanks.

Great explanation & problem! Btw which app you are using? Seems some linux interface.

very good quality question

A quality problem, so much to learn from it.

I suddenly feel much less intimidated by segment trees. I now see that they are just a tool.

This explanation is pure GOLD.

🥵

what's wrong with my code int n, k; cin >> n >> k; string s; cin >> s; int ans = 0; int j = n - 1; while (j >= 0 && s[j] == '0') { j--; } int swaps = n - 1 - j; if (j != n - 1 && swaps <= k) { s[n - 1] = '1'; s[j] = '0'; k -= swaps; } int h = 0; while (h < j && s[h] == '0') { h++; } swaps = h - 0; if (h > 0 && h < j && swaps <= k) { s[0] = '1'; s[h] = '0'; } ans = 0; for (int i = 0; i < n - 1; i++) { int d = (s[i] - '0') * 10 + (s[i + 1] - '0'); ans += d; } cout << ans << endl;

thanks bro umesh

thxx bro

C was really innovative problem

Really good!!

Can you please share the discord server's link?

hey thanks for this 🤟

If you tell the thought process it will even more better

For the case : W=10 , and w={ 4,4,4,8,8} the property is not satisfied.

Thanks for uploading this recording!

Discord server link please?

Loved this explanation

Python doesn't support too deep recursion for this problem, even though I changed the default limit.

Very good explanation. I would like the editorial in codeforces also be this clear. I did not understand why 1 is being subtracted every time . It would have been nice if you have also explained about this in codeforces editorial . Thanks for posting the solution.

Man this what is called an editorial. Every nuisance of the solution is clear. I request to make this case by case analysis of future codeforces round. Really awesome work by IIIT Hyderabad!

it was quite difficult for a C question ,

nice

That's the solution i wanted. Thanks for the video editorial. Please keep uploading these video editorial for the upcoming rounds.it will be very helpful for us.

I have also written a code for this solution but during the contest time it's showing Wrong Answer on Pretest 2 ..... why really don't know ..... it passed all the sample cases given in the question . Can you please tell me what's the wrong there ? If you tell I can share the code with you .

What software are you using to write on the screen? Great questions and explanation btw.

Perfect explanation man Even a noob can understand! I request you to please make vids of contests in general as well. It was just crystal clear!!! Thanks a lot🙏🙏🙏

do i need to study DP, Graphs to be able to solve questions in hashcode?