Bricks and Bags || Codeforces Round 831 Div1+2 Problem C

[@vishalagarwal9392]

absurd explanation of the year!!

[@revanthkalavala1829]

Nice question. I completely went on different path. Nice explanation I Kept min element in bag 3 and maximum element in another bag 2 , and rest all other elements in bag 1 Should have created more examples and tried out but trying out examples in paper, to find optimal answer itself is tiring for this question

[@competitivecoding-newtonsc9601]

Questions like these are quite luck oriented. Either you solve it in 10 minutes or take the whole contest for it.

[@aaravgautam7263]

yep today second option fitted for me😁

[@saumyaranjan9256]

yup, went with the same exact approach for this question and got it wrong.

[@KaifKhan-gs7lw]

I tried the same thing🙃

[@devanshgoel3433]

I also tried the same approach as told by @revanth.

[@praveenkpandey34]

amazing ! editorial , i mean it is way way better then actual editorial ..

[@nikhildixit5687]

such a nice explanation lot of hard work by you, kudos!!!!!!!!

[@leoara8019]

thank you soo much greetings from Syria

[@competitivecoding-newtonsc9601]

No worries 🤠

[@rohitsatpute9578]

What a explanation!! Hats off

[@kamranali4358]

Can you please explain observation 1?

[@competitivecoding-newtonsc9601]

It just means that in the optimal answer, array will be broken into contiguous segments.

[@priovag2632]

Can I ask you why do we use the max for abs(arr[i]-arr[i+1) and abs(arr[i]-arr[i-1]), instead of min? The taker always try to get the closest so the taker can reduce the total score, right? Thank you.

[@sudhanshu9831]

We are assigning the weights in bags. So we'll assign them in a way we get the maximum answer. After that, the taker may choose any way and he'll get the minimum answer as given

[@competitivecoding-newtonsc9601]

This depends on how we assign bags, not on how the taker chooses bricks. So we take max, as it's upto us to assign maximum score.

[@ayyappaswamy8220]

In the first example if we pick bricks whose weights are 1 5 2 the answer would be 7 which is greater than 6. Why shouldn't we pick 1 5 2

[@competitivecoding-newtonsc9601]

Because the person picking the bricks, wants to minimise the score. He will pick the bricks from the given bags such that score is minimum. We want to create these bags such that this minimum score is as big as possible.

[@criticcell]

Hey, I was just wondering about this. I see a lot of channels with the name of Newton School in it. Do they select creators or is this an opportunity open to all. Please specify the details on how someone can be a part of it.

[@competitivecoding-newtonsc9601]

They select creators. I don't know to apply for it though. A senior referred me for it.

[@rohitn6333]

Thanks for this wonderful solution sir could you upload D also

[@competitivecoding-newtonsc9601]

Already uploaded

[@rohitn6333]

@@competitivecoding-newtonsc9601 thank you sir, please continue uploading these solutions. These are very helpful sir Your way of explaining is very good sir

[@xolo2617]

is it type of book allocation problem , where we use binary search ?

[@competitivecoding-newtonsc9601]

Ahh not sure, how you can use binary search on this.

[@kamranali4358]

What could be rating of this problem?

[@competitivecoding-newtonsc9601]

1500

[@kumaramit04-q6c]

I also used the same approach.

[@competitivecoding-newtonsc9601]

Nice ;)

[@CosmoVision1]

Good explanation. However, I put the max element in bag 2, min element in bag 3 and all elements that are left in bag 1 but this solution fails on test 2 and I can't figure out why, could you give an example where this solution fails?

[@competitivecoding-newtonsc9601]

Try 1,2,7,10,10

[@sfg805]

please tell me whats wrong in my approach ll n; cin >> n; vl arr(n); setst; loop(i,n){ cin>>arr[i]; st.insert(arr[i]); } ll ans=0; sort(all(arr)); ans+=abs(arr[0]-arr[n-1]); ll maxi=1e9; ll maxi2=1e9; for(int i=1;i

[@satvikdixit]

Upload D and E as well plz

[@competitivecoding-newtonsc9601]

Will work on it.

[@gurmansingh1878]

Bro can you kindly tell why this code fails...the logic we applied to bag3 (i.e it's size == 1) we can also minimize the size of bag2 == 1 ( we tend to reduce a3) and remaining elements in bag1 void solve() { int n; cin >> n; vll wt(n); for(auto &i : wt) cin >> i; sort(all(wt)); ll ans = (wt.back() - wt.front()); ll m1 = N, m2 = N; for(int i = 1; i < n-1; ++i) { m1 = min(wt[i] - wt.front(), m1); m2 = min(wt.back() - wt[i], m2); debug(m1, m2); } ans += max(m1, m2); cout

[@competitivecoding-newtonsc9601]

How are you placing the bags?

[@gurmansingh1878]

@@competitivecoding-newtonsc9601 by taking ..if bag2 as min then bag3 as max vice-versa ..and just keeping all reamining values i.e from index i = 1 ...n-1 in bag1

[@competitivecoding-newtonsc9601]

What about case like 1 1 2 3 3 ? Isn't it optimal to assign 1 1 or 3 3 to bag 2?

[@gurmansingh1878]

@@competitivecoding-newtonsc9601 Thanks a lot bro...I understand clearly

[@nikhilkumarv2577]

🛐

[@competitivecoding-newtonsc9601]

:)