Someone suggested recording screencasts, I made a poll in my telegram channel (Russian only) and the idea was positively accepted. So, here is my first screencast. I think I will be doing them on every round (if possible) while someone is interested (because it is easy, just 4-5 clicks).

I know that some people would like me to do commentary, but on live rounds I'm trying show my best performance, so it won't happen. But you can suggest some things and maybe I'll do something.

I read Russian problem statements on screencast (because it is more convenient for me), but the code is in C++/Python, so it shouldn't be a problem.

UPD (10.01): I uploaded Hello 2019 screencast and almost uploaded round 530 screencast. I had bad internet connection when I was solving the contest, so I upload only now. Normally I will upload screencasts no more than 30 hours after the contest, so if you are interested in watching it, just go to my channel. I won't make updates like this in the future.

It is this time of year! Many people trying to summarize all the things happened during the year. And since I love to hate people, I want to make a vote for most annoying person on CF in 2018.

I know that I'm in top contributors list for too long, so to vote for given person you should downvote the comment with this person's handle(s). Due to the nature of voting through comments on CF, you can vote for many people, and you can also upvote comments if you find given person's comments helpful/funny/positive.

If you care for your contribution but want to nominate a candidate, you can PM me their handle(s), then I will write a comment to vote. Note that I will write that it was you who nominated this person, so it saves your contribution but not relationships with the nominee. I will ignore messages from people without rating.
If you hate someone already nominated so much that you want them to know, you can also PM me so I'll add you to list of people who nominated given person.
Obviously, you can nominate someone with your own comment, but watch out for downvotes.
Also, you can try to increase your contribution by nominating some very nice people unanimously loved by community. In that case I will nominate you right after that.

I know that I'm obvious candidate, especially after writing this post. But I don't annoy me, so go on and show me your hatred :)

Love you all, Um_nik.

Here are our champions:
-515 rotavirus
-316 riela
-264 MikeMirzayanov
-236 peltorator
-224 CodingKnight
-198 adamant
-166 droptable
-143 kostka
-135 300iq
-135 I_Love_isaf28
-134 I_love_I_love_isaf27

I wanted to write super-harsh blog but then I decided to split it into educational part and whining part. This is whining part. To read some cool stuff go here. This one will also be educational on a slightly different topic. And there will be A LOT of harsh stuff.

I'm not saying that I'm perfect. That round had some really bad problemsetting issues. Problem E was googleable, problem D had very weak tests against very stupid solution (so stupid that none of the authors and testers come up with it, moreover, it is so counter-intuitive that I don't know how so many people independently did it). In two constructive problems participants could look at jury's output, the feature of which I didn't know. And I'm sorry for these issues. BUT. This comment has very negative rating despite being so true. And no one are here to back me up. So I want to speak my truth as I always do.

Let me accompany you to the world of supposedly BAD STATEMENTS. aka statements written by Um_nik.

Problem C from 2nd division of CF Round #518 gets a lot of praise for having very bad statement. Let us look at clarification requests we get on this problem. My answers here are not answers to clarifications from real contest, they are my emotions.

I think that you should be ashamed for your clarifications so I will include handles. This is probably not ethical as a problemsetter to do it. But I don't care anymore. Maybe I will be banned from setting rounds on CF. That looks like a good thing for all of us. You don't want to solve my problems, I don't want to do anything good for you.

"Is it also necessary that if 2 colors harmonize each other they should be connected?" and similar
by sam29, arif.ozturk, aman28rwt, -rs-, Vshining, tgritsaev, L0rdV0ldemort000006, Huvok, seo, shinbay, Apptica, yurics, Distructed_Cat, Adoki, dalbaeb69, chrome, ASUS, cjc, its_ulure, GhostDrag, GhostDrag
Answer: Yes. That's what "if and only if" means. Literally. That is the meaning of these words.

"Excuse me, what is number k in prob C?"
by tgritsaev, fooish, dalbaeb69
Answer: That's a good question, number k is not important for the problem, but I decided to introduce it to ensure that you'll understand that you should place only colored rooks on a chessboard.

"A,B harmonize; B,C harmonize; do A,C Harmonize?"
by harshit601, Fighter.human, Gudleyd, atrophy98, tshr
Answer: No. I have no idea where did you get it from. There is nothing resembling in the statement. Probably it's an attempt to explain some samples, but it is wrong on first sample.

"are there many solutions for sample tests or it's one solution only?"
by Bakry, _QWOiNYUIVMPFSBKLiGSMAP_, bhowmik, mahdi.hasnat
Answer: Another good question, we made global clarification about this later. This was obvious from the samples, I think, but it was a mistake not to write it. I'm sorry.

"In the first sample, we can't go from the blue rook to every green rook!!" and similar
by cfalas, OmarBazaraa, MonsterInMe, DesiChamp, cjc, aman_shahi2, DesiChamp, MonsterInMe, S.Jindal
Answer: Samples are correct. Statement is correct.

"Reply fast please"
by ck98
"The statement looks like it was done with google translate, i don't understand anything"
by AlexArdelean
Answer: That's important information.

"if there is a harmonized pair (a,b). do the ans must contain a set of union of a and b? or i can arrange without any union set."
by Distructed_Cat
"if there is a harmonized pair (a,b). is it a must to make the both set of color a and b connected?" by Distructed_Cat
"And any rook of one color is connected with any other rook of another color that means all rook of 1st color is connected with all other rook of 2nd color if both color harmonise with each other?"
by Fighter.human
"Is it necessary for two rooks of different colors to be in one set?" by bktl1love
"What is harmonize of color??"
by _Muhammad
"can i connect colors other than the described in the input"
by KyRillos_BoshRa
"Specail Judge?"
by C20192413
Answer: Do you fucking read your questions? I like that "I can't solve the problem with operations in statement, can you provide better operations for me?"

Many questions just asking to clarify what the hell is happening in this hellish statement. There are a lot more stupid questions, I just tired.

CF has a large number of participants in each round. If 5% of participants think that his/her time is more important than ours then we will have to answer 5-10 questions every minute. In this round we had 30-40 unanswered questions during an hour, waiting time to get answer to your question is 5-10 minutes. And if we can't understand your question, we spend several minutes on answering. Please. Please. PLEASE. Spend 2-3 minutes trying to answer your question yourself. Use samples. Reread the statement. This is a skill, you have to improve it. Read my blog.

Don't ask "please explain the problem, I don't understand it". We already did explain the problem. The thing called "statement". Read it. Carefully. Why do you think that we will come up with better explanation right now? We prepared this statement several days, we asked testers to read it and say if something is unclear.

Don't say that statement is bad because you can't understand it. The statement can be hard to understand because it requires some knowledge, not because it is badly written. And some statements require to know mathematical notation like "if and only if" in this particular case. I can't come up with better explanation of why there are so many people asking the same question. basically "what if and only if means".

Please assume that authors spend some time on your contest. Much more time than yours 2 hours. Because it is true.

You can remember that once I made round unrated because I found the mistake in authors solution. I didn't wrote a clarification "I think your problem suck" without explanation. I continued my attempts to solve the problem and waited for the end of the contest, then asked what is authors answer for my test. Because I expected that authors did their job well and too busy answering stupid questions during the round.

One more time I spend ~15 minutes trying to prove that my answer for sample is incorrect. When I did it, I send this clarification: "Are you sure that answer for third sample is 34? For me it seems that there are 3 valid trees, in each of them there are 6 possible ways to choose pair of vertices, and for each path there are two possible winning moves for Ember, so the answer should be 36." Compare this to what you do.

I also think that CF should abandon the practice of answering every clarification if possible. It will decrease the number of clarifications and decrease the time to answer one clarification. If I can answer "Read the problem statement", that should be the answer. If I can answer "No comments", that should be the answer.

I think I'm finished with whining about my statements. Now about this doomed community of supposedly smart people.

The thing you did called bullying. There are people who downvote everything I write independent of content. Next there are people who disliked the consequences of my words, even my comment is correct. And there are also people who don't like that I'm saying my thoughts and don't respect their idols.

I liked first comment about "should stop creating problems", and first comment about this after the round. All others don't have own thoughts and have to repeat one thing over and over again. Now it is not a joke, it is being animals.

The same goes for "mind is weak". Yeah, guys, you are so original, it was not me who said it first.

To adequate people in this community: I'm tired of speaking truth. This not your truth? Explain me why am I not right, change my point of view. You think that I'm right in some cases? Speak it up. These animals can attack lone people, but if they will see that I'm not alone, they could think for once. You think that I have no right to insult people, but the idea is correct? Downvote me, say that wording is bad, but express that you agree with meaning.

MikeMirzayanov, I understand that for you quantity is more important than quality. I'm happy that you have your army of stupid fanatics who will write "You forget to hail Mike" under each post. This army has defeated me. Congrats.

But Um_nik, we all know how to read, we have our whooping 2 month of experience! Oh, my sweet summer child, my experiments show that many people with kinda cool achievements like medals on ROI don't know how to read statements. But don't worry, I'll teach you. Well, probably you won't understand anything, because you didn't try to understand anything in your life, you expect all hard work to be done for you by someone else. Let's start!

Basic rules

• The result of reading the statement is usually pure math model. If story helps to build correct understanding, you can keep it, but still try to discard as many unnecessary details as possible.
• Imagine you want to tell the problem to someone else. What parts you want to tell? (According to my PM, this rule won't help you).
• Shorter = better.
• Simpler = better.
• Limitations are part of problem statement. Especially small limitations, because for small data you can try all the possibilities.
• Samples are part of problem statement. After building math model, check that you model working correct on samples, at least on small samples where you can check everything fast.
• Notes are part of problem statement.
• Try to find familiar patterns, maybe objects you already know something about. If you are given some connected graph with exactly one simple path between each pair of vertices, it's called tree. 4 letters instead of 12 words, see?
• Try even harder to spot something strange, something you not expecting. That probably will be the cornerstone of the problem.
• If there is some part of the statement you don't like, try to change that to something you like. Start with understanding the object, then simplify it. There are some problems which can be completely solved by this technique.
• If the model you get is very big, try to split it into some pieces. It would be great if pieces are independent, like 'solve problem1, then use its answer as input to problem2 and so on'.
• On first stages it can be useful to write your new statement. On paper. By hand.
• Problemsetters do not write random things in statements. But why would you believe me, since I'm a bad problemsetter? I don't know, maybe you shouldn't.

Some examples

I'll use Timus for almost all examples. If you see statement in Russian and you don't want to, there is a language switch in upper-left corner.
Assumed workflow: read the statement on Timus, try to understand and simplify it using the rules above. You don't have to read "solution" parts, but I warn you that my concept of reading statements works. I mean, "statement" parts will often contain huge hints to solution.
In most cases I won't write step-by-step explanations how I get the final version. You can say that that's not how teaching works. I can say that you don't want to study. I'll win, because that's my blog.

Networking the "Iset"

Magic Programmer

Scheduled Checking

Work for Robots

Coffee and Buns

Harder examples

Now I'll try to explain something.

GOV-internship 2

Martian Plates

Winnie the Pooh

Titan Ruins: Serial Control

Coolest trick of reading statements

Zamkadye

Different math models will lead you to different solutions

Well, only if you have some patterns of how you should solve problems. And you should have them, patterns are good, they're the fastest way to solve problems.

Empire Strikes Back

Eliminating things you don't like can solve the problem

Last example in this blog will be not from Timus, it is 2016 Yandex.Algorithm Round 3, Problem F, shout-out to Endagorion for creating such pearl.

Problems visible only to participants, so I'll give already simplified problem statement here:

Given a tree, all vertices are initially white. Two players, Red and Blue, take turns coloring white vertices (surprise) red and blue. The game ends when there are no white vertices left. The score of the game is C_R - C_B, where C_R is a number of connected components on red vertices, and C_B is a number of connected components on blue vertices. Red wants to maximize the score, Blue wants to minimize it. Find the score if both play optimally. n ≤ 2·10⁵.

I guess, for international readers all the important information is:

ICPC season 2019-2020, team KAN, Um_nik and someone who we hope to find using this blog.

If you are scared of us and want to skip this season, please do it, we want easy medals, thanks.

CODEFESTIVAL site is updated with info about this year edition but only in Japanese.

Using Google Translate I read that this time only people living in Japan are allowed to participate (at least in the final round).

rng_58, can you confirm this?

Main reason for asking now is that the Finals date is 17.11 which is very close to TCO onsite dates (13.11-16.11).

I want to apologize once more for queue problems. It has also aggravated some tight ML/TL issues which probably would be not so big otherwise. I hope you enjoyed the problems nevertheless.

38799778

38799800 — logs
38799811 — case analisys

38799824

38799831

38799840

I hope that Petr is not mad at me for my joke.

38799850
38799853 — completely different solution with complexity O(6^n / 2)

38799873

38799881

I hope that PrinceOfPersia is not mad at me for my joke.

38799887

Hello!

I'm glad to invite you all to Round 485 which will take place on Tuesday, May 29 at 18:35 UTC+3.

There will be 6 problems in each division, 3 of them are shared between divisions. 2 of div.2 problems created by KAN, 7 others created by me. The contest duration is 2 hours. My problems were originally used in HSE Championship.

I would like to thank Merkurev and GlebsHP with whom I discussed the problems, I_love_Tanya_Romanova for testing, KAN for round coordination and Codeforces and Polygon team for these beautiful platforms. I would also like to thank HSE for organizing the event. Oh, and kb. for writing great legend for one of the problems (he will be surprised).

For those who for some reason like to know score distribution in advance:
div.2: 500 — 1000 — 1250 — 1750 — 2250 — 3000
div.1: 500 — 1000 — 1750 — 2500 — 2750 — 3000
But I didn't properly discuss it with KAN so this is subject to change :)
UPD: Discussed with KAN, scores for C and D in div.2 increased by 250.

Good fun, have problems, read all the luck. As usual.

Oh, and if there will be no bad things, round will be rated. I hope.

I'm very sorry for issues with queue. I understand that it could ruin the contest for many of you. But please be understanding. Bad things happen. It could be some network problems or some electricity problems. MikeMirzayanov, KAN and other people behind Codeforces trying their best to provide good service. But sometimes it's very hard for some reasons. I hope that you enjoyed the problems despite all the bad things.

We decided that round will be rated.

Editorial will be posted tomorrow.

div.1 winners:
1. tourist
2. EvenImage
3. Radewoosh
4. webmaster
5. LHiC

div.2 winners:
1. sminem
2. Fortza_Gabi_Tulba
3. cheburazshka
4. Lomk
5. 2ez4me

Editorial is up.

cgy4ever, when will the round be? clist.by says that it will be tomorrow in 19:00 (MSK), but three days ago it was 18:00 (MSK) and I also cannot find anything related on TopCoder itself. For example, there is no such competition in Web Arena calendar.

Suppose we want to solve a problem by doing binary search on answer. Then the answer will be checked against jury's answer by absolute or relative error (one of them should be smaller then ε). For simplicity we will assume that our answer is always greater than 1 and smaller than B. Because of that, we will always use relative error rather than absolute.

Suppose we have made n iterations of our binary search — what information do we have now? I state that we know that real answer is lying in some segment [x_i, x_i + 1], where 1 = x₀ < x₁ < ... < x_i < ... < x_2ⁿ = B. And what is great — we can choose all x_i except for x₀ and x_2ⁿ.

Now, for simplicity, we will also assume that we will answer x_i + 1 for segment [x_i, x_i + 1] and the real answer was x_i — it is the worst case for us. It is obvious that we will not do that in real life, any other answer would be better, but you will get the idea.

So, what is our relative error? It is . Worst case for us is when relative error is maximal. It is logical to make them equal — exactly what we do by binary search with absolute errors. . We can assume that so . Now we have , but , so . How large should be n to get error less than ε? . Much smaller than .

How to write such binary search? We want to choose m in such a way that or simply .

Now I want to deal with some assumptions I made.

How to choose answer in the end? Again, (it is basically the same as dividing the segment in binary search).

What to do if the answer can be smaller than 1? Try 1; if answer is smaller than 1~--- use standard binary search (because absolute error smaller than relative); is answer is bigger than 1~--- use the binary search above.

P.S. I have never heard about this idea and come up with this while solving 744D - Hongcow Draws a Circle. I'm sorry if it is known for everyone except for me.

If you are registered on Timus, then you should have been received e-mail with invitation to this contest. The contest will be on 19 November in 11 Moscow time.

It was a junior championship but it was 'a little' harder than it was intended. As I remember, Merkurev solved all the problems with only 15 minutes to go, so I think that this contest may be interesting for participants with different levels, up to the strongest ACM teams.

The duration is 5 hours, standard ACM rules. It is a team (up to 3 participants) contest but you are welcome to partcipate individually too.

The authors of the problemset are Um_nik, sivukhin, kb., Tinsane and crassirostris.

Hello!

October Clash on HackerEarth has started and you have 23 more hours to go.

I am the author of this problemset and niyaznigmatul is a tester. I want to thank sivukhin who helped me a lot with preparation.

I hope you find the problems interesting and wish you good luck!

Q1: I'm a newbie. What should I do to become great coder?
A1: Stop doing competetive programming Solve problems.

Q2: I'm doing CP for two months and I'm still not red green. What should I do?
A2: You are lazy and impatient Solve more problems.

Q3: You became a red in less than two years, it is unbelievable!
A3: No, it isn't. You can do it too if you will solve fucking problems.

Q4: You became a red blah-blah-blah such a huge fan blah-blah. Oh, and what should I do to become as great as you?
A4: ... right. You already know the answer. Solve problems. I hate you.

Q5: I'm not good at DP [or something]. What can you suggest?
A5. Maybe you should try to stop asking stupid questions and solve some problems on DP? Or read some blogs and editorials.

Q6: I can't solve a problem / understand your code. Can you help me?
(Well, it is not a bad question in general. It is a good (if you really want me to explain something not to write the solution instead of you) question. But...)
A6: Sure. Can you provide the link to the problem / code?
(I'm not joking, it is a real story).

Bonus!
Q0: Hello bro/sir.
A0: Stop doing this please.

Access to the specified resource has been forbidden.

Codeforces said 20 minutes ago when I was trying to login. The same was about two weeks ago. In both cases I was unable to login for about 10 minutes.
And I constantly logout. Sometimes I can't read 3-4 blog posts before logout.

There were a post about logouting two or three days ago but I can't find it now because search is not working too.

I've noticed during the contest that we can see all the hacks in our room.

Now MikeMirzayanov says that it was a bug and will be fixed soon.

I think it really will be bad if there are tons of hacks. But it is very useful to know which problems can be hacked (I mean which problems was already hacked by someone). Can you add number of successfull and unsuccessfull hacks on each problem to the PROBLEMS page?

I would like to invite you to participate in Ural Regional School Programming Contest.

It is the Ural's qualifying round for All-Russian School Programming Contest and I think that it can be good opportunity for schoolboys to take part in this contest. It will be interesting for students too, I suppose.

Contest starts at 17 October 11:00 MSK at the same time with the onsite contest.

The problemsetters are Um_nik, sivukhin, kb., hx0, Merkurev, KuchumovIlya, droptable (spoiler: it's not about palindromes).

Duration of the contest is 5 hours, ACM ICPC rules. Problem statements will be both in Russian and English. Link to the contest page

569A - Music

Suppose we have downloaded S seconds of the song and press the 'play' button. Let's find how many seconds will be downloaded when we will be forced to play the song once more. . Hence x = qS.

Solution: let's multiply S by q while S < T. The answer is the amount of operations.

Complexity —

569B - Inventory

Let's look at the problem from another side: how many numbers can we leave unchanged to get permutation? It is obvious: these numbers must be from 1 to n and they are must be pairwise distinct. This condition is necessary and sufficient.

This problem can be solved with greedy algorithm. If me meet the number we have never met before and this number is between 1 and n, we will leave this number unchanged. To implement this we can use array where we will mark used numbers.

After that we will look over the array again and allocate numbers that weren't used.

Complexity — O(n).

568A - Primes or Palindromes?

It is known that amount of prime numbers non greater than n is about .

We can also found the amount of palindrome numbers with fixed length k — it is about which is .

Therefore the number of primes asymptotically bigger than the number of palindromic numbers and for every constant A there is an answer. Moreover, for this answer n the next condition hold: . In our case n < 10⁷.

For all numbers smaller than 10⁷ we can check if they are primes (via sieve of Eratosthenes) and/or palindromes (via trivial algorithm or compute reverse number via dynamic approach). Then we can calculate prefix sums (π(n) and rub(n)) and find the answer using linear search.

For A ≤ 42 answer is smaller than 2·10⁶.

Complexity — .

568B - Symmetric and Transitive

Let's find Johnny's mistake. It is all right in his proof except ``If '' part. What if there is no such b for an given a? Then obviously otherwise we'll take b = a.

We can see that our binary relation is some equivalence relation which was expanded by some "empty" elements. For "empty" element a there is no such b that .

Thus we can divide our solution into two parts:

1. Count the number of equivalence relations on sets of size 0, 1, ..., n - 1

2. For every size count the number of ways to expand it with some "empty" elements.

We can define equivalence relation using its equivalence classes.

So first part can be solved using dynamic programming: dp[elems][classes] — the numbers of ways to divide first elems elements to classes equivalence classes. When we handle next element we can send it to one of the existing equivalence classes or we can create new class.

Let's solve second part. Consider set of size m. We have found that there are eq[m] ways to build equivalence relation on this set. We have to add n - m "empty" elements to this set. The number of ways to choose their positions is C_n^k. We can calculate all the binomial coefficients using Pascal's triangle.

So the answer to the problem is .

Complexity — O(n²)

568C - New Language

Suppose we have fixed letters on some positions, how can we check is there a way to select letters on other positions to build a word from the language? The answer is 2-SAT. Let's see: for every position there is two mutually exclusive options (vowel or consonant) and the rules are consequences. Therefore we can do this check in O(n + m) time.

Let's decrease the length of the prefix which will be the same as in s. Then the next letter must be strictly greater but all the next letters can be any. We can iterate over all greater letters and then check if we can made this word the word from the language (via 2-SAT). Once we have found such possibilty we have found the right prefix of the answer. After that we can increase the length of the fixed prefix in a similar way. This solution works in O(nmΣ ) time. We can divide this by Σ simply try not all the letter but only the smallest possible vowel and the smallest possible consonant.

And you should remember about the case when all the letters are vowel (or all the letters are consonant).

Complexity — O(nm)

568D - Sign Posts

Suppose, that solution exist. In case n ≤ k we can put one signpost on each road. In other case let's choose any k + 1 roads. By the Dirichlet's principle there are at least two roads among selected, which have common signpost. Let's simple iterate over all variants with different two roads. After choosing roads a and b, we will remove all roads, intersecting with a and b in common points and reduce k in our problem. This recursive process solves the problem (if solution exist).

Complexity of this solution — . If implement this solution carefully — you will get AC =)

But in case of TL we can add one improvement to our solution. Note, that if we find point, which belongs to k + 1 or more roads, then we must include this point to out answer. For sufficiently large n (for example, if n > 30k²) this point always exist and we can find it using randomize algorithm. If solution exist, probability that two arbitrary roads are intersects in such a point not less than . Because of it, if we 100 times pick two random roads, then with probability such a point will be found and we can decrease k.

All operations better to do in integers.

Complexity — .

568E - Longest Increasing Subsequence

Let's calculate array c: c[len] — minimal number that can complete increasing subsequence of length len. (This is one of the common solution for LIS problem).

Elements of this array are increasing and we can add new element v to processed part of sequence as follows:

1. find such index i that c[i] ≤ v and c[i + 1] ≥ v

2. let c[i + 1] = v

We can process this action in time.

When we handle a gap, we must try to insert all numbers from set b. If we sort elements of b in advance, then we can move with two iterators along arrays b and c and relax all needed values as explained above. This case requires O(n + m) time.

Authors implied solution with O(n) space complexity for answer restoring. We can do this in the following way:

1. Together with array c we will store array c_index[len] — index of element, which complete optimal increasing subsequence of length len. If this subsequence ends in a gap — we will store  - 1.

2. Also, we will store for every not gap — length of LIS(lenLIS[pos]), which ends in this position (this is simply calculating while processing array c) and position(prevIndex[pos]) of previous element in this subsequence (if this elements is gap, we store  - 1)

Now we will start recovery the answer with this information.

While we are working with not gaps — it's all right. We can simply restore LIS with prevIndex[pos] array. The main difficulty lies in processing gaps. If value of prevIndex[pos] in current position equal to  - 1 — we know, that before this elements must be one or more gaps. And we can determine which gaps and what values from b we must put in them as follows:

Let suppose that we stand at position r (and prevIndex[r] =  - 1). Now we want to find such position l (which is not gap), that we can fill exactly lenLIS[r] - lenLIS[l] gaps between l with increasing numbers from interval (a[l]..a[r]). Position l we can simply iterates from r - 1 to 0 and with it calculating gaps between l and r. Check the condition described above we can produce via two binary search query to array b.

Few details:

1. How do we know, that between positions l and r we can fill gaps in such a way, that out answer still the best?
   Let countSkip(l, r) — count gaps on interval (l..r), countBetween(x, y) — count different numbers from set b, lying in the range (x..y).
   Then, positions l and r are good only if lenLIS[r] - lenLIS[l] = min(countSkip(l, r), countBetween(a[l], a[r])). countSkip we can calculate while iterates position l, countBetween(x, y) = max(0, lower_bound(b, y) - upper_bound(b, x)).

2. What to do, is LIS ends or begins in gaps?
   This case we can solve by simply adding  - ∞ and  + ∞ in begin and end of out array.

Complexity — . Memory — O(n + m).