Codeforces Round 865 (Div. 2) B-E / (Div. 1) A-C (video editorial)

[@wesleyso0]

I love your explanation for C. Super easy to understand! Thank you for your help. Keep on making these videos!

[@1732ashish]

thank you for making these video, although I am low in rank, this is still helpful to me.

[@DamianKatershorts]

Awsome waiting for more

[@harrypotteredits7265]

Thanks for the video waiting for more

[@ANANDKUMAR-jk9yp]

E was a very nice one!

[@shuvo_14]

GReat effort! Take love and keep doing this

[@aviralarpan7350]

This is the best codeforces contest explanation i have seen till now

[@shubhamkumar3198]

Thank you so much for your wonderful explanation keep posting these solution it is very helpful

[@aviralarpan7350]

Wow, thanks a lot, really great explanation, if possible, please continue for the upcoming div2 rounds

[@info_ledge1473]

your explanation is awesome

[@ash5682]

Thank you!

[@kabboghosh1853]

I didn't solve in night i always know I need ur help , waiting for ur video

[@culverkwan9834]

The problems that you explained on have pairwise distinct authors. Guess who authored which!

[@rembocoder]

It's impossible to guess.

[@galihmahotama888]

why I have to sum of those differences on even positions? can I count the sum of those differences on odd positions and if the sum is >= 0 then the output is YES, opposite the output will be NO

[@rembocoder]

No. Why is explained in the video.

[@VasheshJ]

Hey, For Div2 - C, I tried the following approach and am not able to figure out it why it is giving a WA. Could you help me figure out any case it might be failing? def solve(n, array): maximum = 10 ** 9 + 1 for i in range(1, n): if array[i] < array[i - 1]: if i + 1 < n: diff = array[i - 1] - array[i] array[i + 1] += diff array[i] += diff else: print("NO") return diff = array[i-1] + maximum array[i-1] -= diff array[i] -= diff print("YES")

[@rembocoder]

The real question is why should it work? Do you have a proof?

[@VasheshJ]

@@rembocoder I think it is a greedy idea of making all the initial values as minimum as possible. You can proof that all the values except the last two can be made non decreasing. Now we can just check for the last two values of the array to see if they are non decreasing or not and give the answer. Upd: somebody helped me figure out the soln…It was falling for the test case 1 7 1 1000000000 1 100000000 1 1000000000 1 I increased the maximum value by multiple of 100 and it got AC. I think the idea for the minimum value depends on how big the array values are.

[@VasheshJ]

@@rembocoder Thanks anyways