Maybe, they call speedforces because the difficulty level in between two consecutive problem is too high. The person having more skills/practice got beaten by the speed of some other participants just because he was unlucky in starting problems in some of the contests. And anyway both candidates won't be able to solve the next problem because it's too difficult that results in negative rating change. <- My perspective :)

Codeforces problemset section

can anyone please give hints for problem E.

Any update regarding mathura regionals?? And where will the mathura regional be held (mathura or kanpur)??

What's the last date of registration for the three regionals??

Maybe Hello 2023 will be enough, if you get a rank of 500-600.

Best of luck.

Last chance to see myself as CANDIDATE MASTER Adarsh8409 on NEW YEAR EVE.

Is there any possibility to get the editorial for these problems. We would be very thankful to you.

Btw, problems were really interesting :)

Problem- E:- Can anyone tell, which case they where missing while they where getting WA at test case 16.

expected: '1476747', found: '996573'

How to correct it.

Thanks in advance :)

Two contest in one day with gap of only 30 minutes. Can authors postpone it for 1 day.

I think, for every contest we should try to solve problems till end of the contest. Sometimes, last problem gets accepted in last minute of the contest.

If the contest gets postponed, it will be good for everyone.

Hello authors, Can you please postpone the contest by half an hour as Google kickstart will end at 20:30 IST. We will be very very grateful to you.

you can split the array in 2 parts every time. and you have to check that the elements in first part is always less than second part for every subtree. if not, increase the count and repeat the same process till the leaf of the tree.

Use recursion for solving the problem.

You can take reference from my code-> https://mirror.codeforces.com/contest/1741/submission/175603313

Problem E was a perfect 1D dynamic programming question and the best problem for div3 E.

Overall Great questions. Congrats to authors.

Yes you are right

Maybe I am unable to make them understand my solution :')

I think D was not too hard. It will be of rating 1600 approx.

It can be solved by simple observation, that are:-

1) If the no. of ones are odd then answer is -1.

2) Simply divide the array in two subsequences, as per your choice. I personally have divided the array in two subsequences on the basis of parity (i.e. odd and even index)

3) Now, see if the ith index of subsequence (i.e. 2*i and 2*i+1 index of original array) are same then we can just ignore that. else you can store these pairs of index in some temporary vector.

4) You will notice that the size of that temporary array is even (because statement (1) is false).

5) Now for every "two" pair of the vector choose one '0'th index form pair 1 and '1'th index from pair 2.

6) You can notice that after the operation of right shift, all the bits will flip as the size of the array will even always. Ex- 01 01 01 -> 10 10 10.

7) In short, every two pair in temporary vector will satisfy there requirements.

8) example- Lets say pair1(1,0) and pair2(1,0) you take 0'th from pair1 and 1'th from pair2. and apply the operation then the pairs will become pair1(1,1) and pair2(0,0) and that's what we need. "NOTE- Store index in the pairs." I have used 0 or 1 for better understanding.

8) And we get our desired answer.

You can take some reference from my code :) My submission id- https://mirror.codeforces.com/contest/1736/submission/175419375

Thanks for reading it, hope this would help.

Happy coding :)

like there was questions from implementation, STL, graph, DP. One can learn new concepts from these question.

Overall contest have good mixture of all concepts of competitive programming. Nice contest @low_ sir.

Solved problem A in 6 minutes. it was quite easy for me.

Its my one of the best contest ever. Thanks codeforces:)

What is the criteria of random distribution of the goodies :)

I am eagerly waiting for the first "random" merchandise packages.

Nice explanation brother :)