We invite you to participate in CodeChef’s Starters 11 this Wednesday, September 15th, rated for both division 2 & 3 participants.

Time: 8 PM — 11 PM IST.

Joining me on the problem setting panel are:

Setters: Utkarsh Utkarsh.25dec Gupta, Soumyadeep palsoumyadeep Pal, Aryan Agarwala, Lavish lavish315 Gupta, Akash Whiplash99 Bhalotia, AbdelAziz Tsetso Mostafa, Mradul bhatnagar.mradul Bhatnagar, Surya evilbuggy Prakash

Tester: Samarth gupta_samarth Gupta

Statement Verifier: Sarthak SarthakAjmera Ajmera

Editorialist: Taranpreet Discombobulated Singh

Contest Admins: Utkarsh Utkarsh.25dec Gupta, Daanish Mahajan

Video Editorialists: Chirayu Chirayu Jain, Prachi agarwal19 Agarwal, Darshan darshancool25 Lokhande, Yashodhan Molotov Agnihotri, Shivam Bohra, Aryan Agarwala, Meet mithh_gemphir Singh Gambhir, Rajarshi RestingRajarshi Basu, Bharat Singla, Rishab Nahar

The video editorials of the problems will be available on our YouTube channel as soon as the contest ends. Subscribe to get notifications about our new editorials.

Also, if you have some original and engaging problem ideas, and you’re interested in them being used in CodeChef's contests, you can share them here.

Hope to see you participating.

Good Luck!

How many problems would be there for each division?

6 in division 2 and 7 in division 3. Both divisions sharing 5 problems

How to solve ATM Queue?

What was there in the explanation of ATM queue?? if 2 is in front of 4, why can't 4 overtake?? It is written the A2 > A4 but A2=2 and A4=4. How is this explanation valid???

2 and 4 , they are talking in terms of indices and not power values! So, A2>A4 (3>1).

UPD: Never mind got it

Basically , in the t(th) second you can think that the following code is executed: pop the a[t]. for e from t+2 to n: if a[e]>a[e-1] swap (a[e],a[e-1])

Because first 4 will overtake 2 and then 5 will also overtake 2 in the same second

How to solve COUNTA? there's no editorial for it

Currently the official editorial is yet to be published. I can provide you a rough soln outline.

B[i] = min(A[i] , A[i+1])

let us calculate dp[N][2], where dp[i][0] represents B[i] = A[i], and A[i+1] > A[i] and dp[i][1] represents B[i] = A[i+1], and A[i] >= A[i+1].

We can now calculate dp[i+1][0] and dp[i+1][1] using dp[i][0] and dp[i][1], and making cases on A[i] and A[i+1]

Think along this direction

The Div 2. problemset had really nice problems. Thank you for the nice problemset! It more than made up for the extremely weak testcases in problem C.

Thanks for your feedback. Sorry about the weak test cases in C. Will not repeat this in future contests.