Kuroni's blog

Codeforces Round #616 Editorial
By Kuroni, history, 5 years ago, In English

Hello everyone, this is the editorial for Codeforces Round 616 (Div. 1) and Codeforces Round 616 (Div. 2)! Along with the solution to each problem, we will have the theme and easter egg solution as well! I hope you all enjoyed our problems ( ´ ▽ ` )b

1291A - Even But Not Even

Author: 265918

Tutorial
Implementation

1291B - Array Sharpening

Author: hugopm

Tutorial
Implementation

1290A - Mind Control

Author: Ari

Tutorial
Implementation (quadratic)
Implementation (linear)

1290B - Irreducible Anagrams

Author: Ari

Tutorial
Implementation

1290C - Prefix Enlightenment

Author: hugopm

Tutorial
Implementation (preprocess with DFS)
Implementation (dynamic bipartite DSU)

1290D - Coffee Varieties (hard version)

Author: hugopm

Tutorial
Implementation

1290E - Cartesian Tree

Author: gamegame

Tutorial
Implementation

1290F - Making Shapes

Author: Kuroni

Tutorial
Implementation

Theme and easter eggs

Spoilers
  • Vote: I like it
  • +239
  • Vote: I do not like it

Comments Comments (119)
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5 years ago, # |
  Vote: I like it +156 Vote: I do not like it

Thanks for the easter eggs, now my next few nights would be sleepless. :)

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5 years ago, # |
  Vote: I like it +13 Vote: I do not like it

can you please share the test case generator for Div2-E problem?

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    5 years ago, # ^ |
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    It's easy. Just assign every number from $$$1$$$ to $$$n$$$ at most 2 different numbers from $$$1$$$ to $$$k$$$ , which means which set the number should belong to.

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5 years ago, # |
  Vote: I like it +26 Vote: I do not like it

One of the best contest it was !

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5 years ago, # |
← Rev. 3   Vote: I like it +8 Vote: I do not like it

Nice problems and fast AF editorial and good pretests.

Thanks man. :)

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5 years ago, # |
  Vote: I like it +28 Vote: I do not like it

WOW! VERY quick editorial! Thanks)

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5 years ago, # |
  Vote: I like it +31 Vote: I do not like it

thanks for the fast editorial

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5 years ago, # |
  Vote: I like it +13 Vote: I do not like it

Thank You very much for the problems.Learnt to think for basic things first in easy problems rather than taking cases and complicating.

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5 years ago, # |
  Vote: I like it +58 Vote: I do not like it

Thank you for samples in C.

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5 years ago, # |
  Vote: I like it +14 Vote: I do not like it

Div1 BCD are really good, thanks!

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5 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Thanks for the fast editorial.

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5 years ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

Doubt For Mind Control Problem 1290A :->

In the editorial it is mentioned that iterate over y. But why ?

Say my k = 3 and m = 6 for some n then say i force 1 guy to take first element and rest 2 to take last. Now the three people i have no control over can choose first element or not.

i.e i cannot guarantee that the next three elements (after the first one is chosen by the guy i forced) will be my element all the time. but i can say for sure is that:

the element at fourth index i.e after 3 elements can be my answer. This will of course be compared with the number obtained from the end.

so basically for each person i force to pick firt element say i and y ppl i cannot force. The only element i can pick is (i + y + 1) from the start. We will get one answer from last as well.

So should i just iterate over i from 0 to k.

calculate the expected number from first and last. Take their minimum

compare globalmaximum with minimum obtained

print globalmaximum

Plz Help

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    5 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    I'm not so clear about your doubt, so here let me try to explain one insight without $$$ y $$$ ... Observe the array below

    $$$[2, 9, 2, 3, 8, 5]$$$

    Consider when

    $$$k=0, m=4$$$

    All possible final states will be like these sub-arries:

    $$$[2, 9, 2], [9, 2, 3], [2, 3, 8], [3, 8, 5] $$$

    For every states, the answer is:

    $$$ [2], [9], [8], [5] $$$

    So you can find in the worst situation, the player will get 2 as his final number.

    Now consider how does "k force times" affect the answer. Actually it you can force some opponents to avoid some prefix or suffix states. (you can't avoid one state in the middle, sorry I can't explain this point.)

    In the case above if k=2, you force 2 opponents to take the "2" in the front and the "5" in the back.

    Then all possible final states will be like these sub-arries:

    $$$ [9, 2, 3], [2, 3, 8] $$$

    For every states, the answer is:

    $$$ [9], [8] $$$

    After this force, the worst anwser gets greater, right?

    The focus is to find the globalmaximum for every possible states' answer.

    Hope this can help you understand this solution better.

    Nice round anyway. lol

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      5 years ago, # ^ |
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      Thank you.

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      5 years ago, # ^ |
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      How can we say that we can only force some opponents to take first element? I am really confused with this question please help although your explanation was very good but this part that i mentioned is yet not clear.

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        5 years ago, # ^ |
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        Maybe this means in some stage of the whole process, you are allow to force someone make specific choice. Like in a case when the i-th opponent take the front might make the lowest answer worse (leave some small numbers in the back), then you force him to take smaller number in the the back. Could this help?

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      4 years ago, # ^ |
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      when m=2 and my position=4,why the answer for [2,9,2,3,8,5] isn't always 9? because if 1st person take 2 ,i will force 2nd and 3rd person to take last element. and if 1st person take last element,i will force 2nd person to take 1st element and 3rd person to take last element. thus i always get 9.

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        4 years ago, # ^ |
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        what if first person take 5? because you cannot bound first person to take only 2 if he takes 5 then answer changes and you have to tell the maximum of worst case.

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5 years ago, # |
  Vote: I like it +7 Vote: I do not like it

did anyone else solved c with binary search?

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    5 years ago, # ^ |
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    Me, but it seems unnecessary. :(

    I rather overthought.

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5 years ago, # |
  Vote: I like it -8 Vote: I do not like it

I learned a lot from Div.1 C even though I was not able to solve it during the contest. Thank you for making such good problems!

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    5 years ago, # ^ |
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    bro people toxic af u didn't even say anything bad and -8

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5 years ago, # |
  Vote: I like it +28 Vote: I do not like it

The six digit codes...how are they more than just random 6 digit numbers? What makes them an easter egg?

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5 years ago, # |
  Vote: I like it -8 Vote: I do not like it

For Div. 2 B. I consider the following:

Just consider all indices that contain the max value in the array. Then, for each, try to find if all the elements to its left can be made strictly decreasing. Then do similarly with the elements to its right: try to make them strictly decreasing.

If you could find one such index then the answer is true. Why do I get WA here...?

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    5 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Take $$$[1, 2, 1, 3]$$$ for example. Your algorithm will give NO, but a possible sharpened array generated from this could be $$$[1, 2, 1, 0]$$$.

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      5 years ago, # ^ |
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      Thanks! I got obfuscated believing that if there's a YES answer, it MUST be when a[k] = MAX(a) :-(

      Good lesson!

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5 years ago, # |
  Vote: I like it +13 Vote: I do not like it

Thanks for the fast editorial and an interesting D!

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5 years ago, # |
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For div-2 D,I think atleast two different character is sufficient instead of three.. Pls give any counter example.

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5 years ago, # |
← Rev. 2   Vote: I like it +4 Vote: I do not like it

can some one explain how monotonic dequeue has been used in Mind Control (Div2 C / Div1 A) to solve the problem in O(n).

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    5 years ago, # ^ |
    ← Rev. 3   Vote: I like it +8 Vote: I do not like it

    First, calculate b[] in O(n)

    Then, use a min queue. Start by pushing the first m-k values into the queue, and define ans = 0.

    Then perform the following steps k+1 times:

    1. Query v = min in queue, and set ans = max(ans, v)

    2. Add the next value of b into the queue

    3. Pop a value off the front of the queue

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5 years ago, # |
← Rev. 2   Vote: I like it -21 Vote: I do not like it

ignore this comment, it is wrong.

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    5 years ago, # ^ |
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    This wont work. They do not take maximum, the take that ends that the max of the remaining two elements will be minimum.

    Same true for your k choices, you do not take minimum, you take that what maximizes the possible result at the end.

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      5 years ago, # ^ |
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      yeah, it is wrong, just find a simple counter

      but for the remaining M — K people isn't it optimal to take max? because if no one take it, i will take it in the end and the answer will better for me

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        5 years ago, # ^ |
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        If you take max, the next element could be even bigger. So, what counts is what is left after you take them, not what you take.

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5 years ago, # |
  Vote: I like it +36 Vote: I do not like it

Waiting for the rating update. Wish to see blue tag in my own profile for the very first time. Thanx to the whole CF community for making me this kind of eager for competitive programming.

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    5 years ago, # ^ |
      Vote: I like it +26 Vote: I do not like it

    And now you have it! Congratulations!!!

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    5 years ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    Congratulations !

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    5 years ago, # ^ |
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    Actually I won't be able to explain, how much happy I have become right now.

    Once it was just a dream for me. But it just became true.

    Wish to become PINK soon. Pray for me. Thanks to the whole community.

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    5 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    Congratulations !

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5 years ago, # |
  Vote: I like it +13 Vote: I do not like it

Binary search tag D?!

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    5 years ago, # ^ |
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    To know how many distinct characters in range [L,R], you can for each character do a binary search to get the first occurrence for this character greater than or equal to L,if it was smaller than R then you know that this character exists in the range .

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    5 years ago, # ^ |
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    Sure, you can store the position of each time a character appears on the string and find the number of times the character $$$c$$$ appears in the range $$$[l,r]$$$ as $$$upperbound_c(r) - lowerbound_c(l)$$$. That technique could be useful in other problems. You can check this code to know how to implement it.

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5 years ago, # |
  Vote: I like it +36 Vote: I do not like it

tourist back to 1. Finally bug on codeforces resolved.

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5 years ago, # |
← Rev. 3   Vote: I like it -28 Vote: I do not like it

.

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5 years ago, # |
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From 1290C Sol: "Since the answer exists for i=n, there exists a such partition of the graph (into "red" and "blue" nodes). We can find it with usual dfs, and keep it for lower values of i"

Why is this coloring still optimal for i < n? What if there is a more optimal coloring? Can someone give me a proof for why we can still use the same coloring for lower values of i?

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    5 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    From the full graph to the graph representing the state $$$i<n$$$, we do nothing but take away edges. That means intuitively, there are less constraints to force the colorings, so the coloring still works.

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5 years ago, # |
  Vote: I like it -15 Vote: I do not like it

Are there some strong folks who can explain how the "Cartesian" tree from div1 E actually constucted. Preferably, explanation of the first example, sequences of length 4 and more.

What is the step of recursion. Why for length 4 from example numbers from left and right parts of MAX are mixed. Can't grasp at all.

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5 years ago, # |
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why do we do strategyAns = min(strategyAns, caseAns); and not max pls help(Div2C)

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    5 years ago, # ^ |
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    because we want to take the maximum minimum of the numbers left. The first sample case was well explained: 

    2 9 2 3 8 5
The first one was forced to take 5 
2 9 2 3 8
The second one was forced to take 2
9 2 3 8
and One guy left, he could take 8 so we can take 9, but if he took 9 then we can take 8. so we just took the maximum minimum, and that is the least maximum number that we could take.
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5 years ago, # |
  Vote: I like it +66 Vote: I do not like it

I think I have a bit simpler implementation for Div1 C.

I use an ordinary DSU, with representing each original node $$$x$$$ with two nodes, one that represents that $$$x$$$ will be chosen and the other one representing that $$$x$$$ will not be chosen, let's call them $$$x_{true}$$$ and $$$x_{false}$$$.

Also, each root has a cost, which is the cost to choose that root. Initially, the cost of each root is $$$1$$$ if it is a $$$true$$$ node and $$$0$$$ otherwise.

I also create a dummy node representing the $$$nochoice$$$. The cost of its $$$true$$$ node is $$$0$$$, and the cost of its $$$false$$$ node is $$$OO$$$.

To merge two nodes $$$x$$$ and $$$y$$$, if they must take the same value, then $$$join(x_{true}, y_{true})$$$ and $$$join(x_{false}, y_{false})$$$, else $$$join(x_{true}, y_{false})$$$ and $$$join(x_{false}, y_{true})$$$.

To force a node to be $$$true$$$ or $$$false$$$, then join it with the $$$nochoice$$$ node.

The solution maintains the total cost and updates it upon merging in a similar manner to the tutorial's solution. The dummy node makes it unnecessary to care for overflows or minimize the costs with $$$OO$$$, since that $$$OO$$$ will only be counted once, in the dummy node.

Code: 70098404

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    5 years ago, # ^ |
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    Had similar idea, the approach is kind of a 2-SAT (Boolean Satisfiability)

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5 years ago, # |
  Vote: I like it -8 Vote: I do not like it

Thanks for the useful tutorial <3

Have a nice day <3
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5 years ago, # |
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Regarding D,

In the problem: partition the graph having directed edges $$$(i, j)$$$ for all $$$i < j$$$ into edge-disjoint paths,

Isn't it possible to prove using Hall's theorem that number of disjoint paths must be $$$ \geq $$$ about $$$ \frac{n^2}{4} $$$?

This means that $$$1.5$$$ is the best possible factor for the above approach. How do you get a factor of $$$1.2$$$ in randomized DFS then?

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    5 years ago, # ^ |
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    The graph is not directed. As long as we visit vertices $$$i$$$ and $$$j$$$ consecutively in a path in any order we will be able to remove all equalities. The only difference is which of the blocks will have elements deleted.

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5 years ago, # |
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Thanks for the fast editorial.

Is here a guy who knows a different solution of Irreducible Anagrams?

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    5 years ago, # ^ |
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    I interested on it because Irreducible Anagrams has binary search, data structures, strings, two pointers problem tags .

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      5 years ago, # ^ |
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      There are quite a few other solutions to Irreducible Anagrams, differing on how the third condition is handled. I believe the one in the editorial is the most straightforward, but some other possible solutions we intended to pass include:

      • Find for each left endpoint $$$l$$$ the maximum $$$x$$$ such that $$$s[l, x]$$$ contains at most two different characters. After preprocessing this allows us to answer queries in $$$O(1)$$$. This can be done in $$$O(n)$$$ using a somewhat straight forward two pointers algorithm, for a final complexity of $$$O(n + q)$$$.

      • One can use any of the standard solutions for the classic "count number of distinct values in a range" problem, such as Mo's Algorithm in $$$O(n\sqrt{q})$$$, or Sorting + Fenwick Tree in $$$O(q \log (n))$$$.

      • Some other silly solutions also passed, such as using segment trees instead of prefix sums to find whether a substring contains a certain character, which results in $$$O(26 \log n)$$$ per query. This is a rather strange thing to do, but some people did actually pass with solutions like this ¯\_(ツ)_/¯.

      I'm sure there's many other solutions that could get accepted in this problem, as constraints were low enough to allow basically everything that isn't straight up quadratic to pass.

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        5 years ago, # ^ |
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        Your solution is understandable readable. thanks ;)

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5 years ago, # |
← Rev. 2   Vote: I like it +8 Vote: I do not like it

Nice and clear problems with good pretests. Liked it very much!! :)

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5 years ago, # |
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Thanks for the contest (And I won't definitely sleep a few nights)!

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5 years ago, # |
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Eurrghhhh... I just realize what the easter eggs mean...

I'd never trust weebs ever again >: (

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5 years ago, # |
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hard version of question Div2-C is when you can chose from all numbers(not just from end or front) and your friends must chose from front or end.

i misunderstood and spent about two hours to solve it and after solving i got that i misunderstood

however i solved question C and learnt many thing

one of them is check test-cases before solving question

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    5 years ago, # ^ |
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    I don't fully understand how this is harder, tbh? let dp[i][j][k] be the answer for the subarray from i to j with k people for you to manipulate. Then dp[i][j][k] = max(dp[i+1][j][k-1], dp[i][j-1][k-1]). When k = 0, suppose there are p people left to pick who you cannot manipulate. Then you know that they will leave some contiguous subarray of length (j — i + 1 — p). You can iterate over all subarrays of such length and pick the minimum of the maximum of all subarrays.

    I did exactly this except instead of picking the min of max of all subarrays, i picked the min of max of all (first element/last element) of subarrays. 70060565

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5 years ago, # |
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So fast

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5 years ago, # |
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I tried to understand the statement of Div1E (Cartesian Tree), but failed. If someone understood and can clarify one thing in construction of the tree, i will be very grateful.

In the very first example, given the sequence 4, 2, 7, 3, 5, 6, 1. According to algorithm, we should take maximum, 7 in this case, so position x = 3(is it correct?). Then we construct trees for [4, 2] (left tree) and [3, 5, 6, 1] (right tree) (again, any missunderstand?). Then, due to step 5, left and right constructed trees become left and right subtrees and the root is temporary removed number. And it maximum, 7 in this case. But in example the resulting tree has root that is not 7 and 1 and 2 in the same subtree, while they were in different parts after breaking the sequence.

Where am I wrong?

P. S. I see that it is ordered as BST, and what i do is a mess. That's why i ask for some help.

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    5 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    The announcment for problem E says: "In the notes to samples, the nodes in the tree are labeled by indices, while the tree in the explanation is labeled by value. Sorry for the inconvenience caused. The problem doesn't change."

    That's why the root of sample tree isn't "7". It labeled "3" because 7 is third element of the array (x = 3). Also 1 and 2 are in the same subtree due to they're indicies of 4 and 2 (a[1] = 4 and a[2] = 2).

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      5 years ago, # ^ |
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      Thaks a lot! Short and clear explanation. Finally I can enjoy the beaty!

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5 years ago, # |
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I want to know that if the problem C has a better solution.

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    5 years ago, # ^ |
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    What kind of "better" do you want. We have a linear solution already.

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5 years ago, # |
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Div.2 B. What is test case 2: 109th line? It seems I'm wrong with my solution, but I can't hack myself. I prepare pyramide-like triangle of "minimal possible values", and then check if all values of an array are higher (70047658).

UPD. Solved. My "pyramide" should be sharper at a top in case of even number of elements!

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    5 years ago, # ^ |
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    did almost same and getting the same problem

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5 years ago, # |
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whats wrong in my code i check for min number at respective indices of array for n = 4 it should be 0 1 2 0 or 0 2 1 0 and similarly for other n values (problem 2 )

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    5 years ago, # ^ |
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    Thanks. I found my mistake by looking at your examples.

    Ok. How do you expand "pyramide" of higher N? E.g. n == 6.

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      5 years ago, # ^ |
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      0 1 2 3 1 0 or 0 1 3 2 1 0 this is how

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5 years ago, # |
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Thanks for the problems, I really enjoyed them. Here's an alternate solution for C.

We will find the answer for each $$$i$$$ by iterating from $$$1$$$ to $$$n$$$. For each set, we associate a cost of picking that set. Initially this cost is $$$1$$$ for every set.

If $$$i$$$ is currently off, then we pick the lowest cost set that contains $$$i$$$. We are going to keep transforming the sets and costs so that doing this produces the correct answer.

If $$$i$$$ is off, then in this and all future iterations exactly one of the sets that contains $$$i$$$ should be selected. If there is only one set containing $$$i$$$, we remove it. If there are two, say $$$S$$$ and $$$T$$$ where $$$S$$$ was selected in the current iteration, then we want to have the ability to unselect $$$S$$$ and select $$$T$$$ instead in later iterations. The set of positions that get flipped by doing this is $$$U = S \text{ xor } T$$$ and the cost of doing this is $$$cost(T) - cost(S)$$$. So we can enforce this by removing $$$S$$$ and $$$T$$$ and adding this new set $$$U$$$ with this cost. (this is kind of similar to residual edges in Ford-Fulkerson; here U is a residual set)

If position $$$i$$$ is already on, then if it is contained in exactly one set, this set should never be picked hereafter, so just remove it. If it is contained in two sets $$$S$$$ and $$$T$$$, then in all future iterations either they should both be picked or neither. We can accomplish this by removing $$$S$$$ and $$$T$$$ and adding $$$U = S \text{ xor } T$$$ with cost $$$cost(S) + cost(T)$$$.

70094066

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5 years ago, # |
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what is the reason behind comparing the vallues with array indexes?? in B cant get this:)

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    5 years ago, # ^ |
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    Because the array wont be strictly increasing if the values of array are less than the sequence 0,1,2,3,..,n that is similar to array indexes.

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5 years ago, # |
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Can someone explain the div2.E's the way to maintain the number of the red and blue nodes,I can not understand the code's defining of l,r...

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5 years ago, # |
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I have a doubt in Div1A/Div2C: Why should I select the control the first $$$k$$$ persons in the queue, and not in some other order: like why not first $$$m-1-k$$$ can select randomly, and the last $$$k$$$ elements should select the element I forced them too. Please help.

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    5 years ago, # ^ |
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    we will choose the first K cause we want the answer for the person in the mth position to be maximum after he chooses he will leave so the people after him will have no effect on the value X but I didn't understand exactly the intuition behind how to get the optimal answer

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      5 years ago, # ^ |
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      Please don't answer questions you don't understand: I am asking about which $$$k$$$ people to choose for controlling.

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    5 years ago, # ^ |
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    I have the same question. How to choose which k person to control? The first k? Or others?

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    5 years ago, # ^ |
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    Basically, in order to know what element that $$$m-$$$th person picks, you need to know after $$$m-1$$$ persons picks the elements themselves, what the remaining subarray is. And the remaining subarray only depends on the number of persons choosing the head element (or equivalently, the amount choosing the tail element) in $$$m-1$$$ persons. It indicates that the order is not important, and you can pick any $$$k$$$ persons to control.

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5 years ago, # |
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i didn't understand div 2 C solution can anyone explain it in a more intuitive level thanks in advance

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    5 years ago, # ^ |
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    Brute force on the persuade people by taking two-pointer one starting from 0 index(say left) and one at n-min(k,m-1)-1 (say right), then for each pair of left and right, find the minimum element you can get by brute-forcing from left to right with the same approach of taking two-pointer one starting from left(say l) and another at n-m+left(say r), then find minimum from max(a[l], a[r], r till right.

    Do the same for all the values of left and right. And the maximum of all the values will be the answer.

    70354436

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5 years ago, # |
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Can you help me to find the time complexity of my submission with explanation for Div2 C problem 70354436

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5 years ago, # |
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Can someone please explain me how is the color of the each node is being maintained in div 1 C. I tried to understand it so much but I couldn't. Is it some kind of common trick that I am unaware of? val[v.fi].xo = 1; What is this line actually doing?

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    5 years ago, # ^ |
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    That is maintaining the parity equality between u and dsu[u]. If u and dsu[u] are colored with the same color, val[v.fi].xo = 0, else val[v.fi].xo = 1.

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5 years ago, # |
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[deleted]

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4 years ago, # |
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What is wrong with my solution ? I m not getting the case where my solution fails So please help me

DIV2/C

solution link : https://codeforces.me/contest/1291/submission/82024558

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4 years ago, # |
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Why can't we assign our k(effective) friends after rest random friends. I think we can get better number if we do that . for eg:

6 4 2

2 9 2 3 8 5

here, 1 -> random and 2 -> friend [9 2 3 8 5] or [2 9 2 3 8] and we get 9 as our answer.

correct me if I'm wrong.

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    4 years ago, # ^ |
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    You have to choose what to do with the people you control before the process starts. In this case, what you do with the 2 controlled people is different depending on what the first person does; that's not allowed.

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4 years ago, # |
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I have spent days on Div2C. I have read all the comments. I dont get either the editorial or any comment's approach. If anyone can explain it in simple words, Please help me!

I have to say this, "This is one of the most unclear Editorials I have ever seen."

Here's my TLE code for a very simple and direct brute force I tried.

This is recursion based. If I have any k type people left, I take them in 2 ways:

-> force him to select first

->force him to select second.

Then I take maximum of these above 2 case.

If I have k=0 left, Then for the remaining m, We check all possible cases and find out the minimum. (I believe my solution is correct, but gives TLE.)

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    4 years ago, # ^ |
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    Yes, your solution seems to be correct for me. But i think the reason for TLE is it's exponential complexity.

    I will explain you my approach -

    let 

    6 4 2
2 9 2 3 8 5

    n = 6, m = 4; k = 2

    first loop take care of number of controlled friends in the left ( 0 to k)

    second loop take number of random in the left( starting from zero to (m-1-k) )

    and then we can take minimum of available left and right number in first loop and finally take maximum in first loop

    for eg:

    1)
2 9 2 3
2 9 2 or 9 2 3
min( max(3, 9), max(2, 2) ) --> 2

2)
9 2 3 8
9 2 3 or 2, 3, 8
min( max(9, 3), max(8, 2) ) --> 8

3)
2 3 8 5
2 3 8 or 3 8 5
min( max(2, 8), max(3, 5) ) --> 5

maximum in all three is 8.

    Link to my code

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    4 years ago, # ^ |
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    I also have read editorial and comments after solving this with dp but have not got anything clearly . And your solution is correct . Just memorize the answers of each state . Here is my Dp solution 87047732

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    4 years ago, # ^ |
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    Thanks! Got it!

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4 years ago, # |
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This problem can also be solved by dp (which is not in the tag section), if anyone is interested, here's my solution: Code

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4 years ago, # |
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Didn't knew codeforces had people of culture ( ͡° ͜ʖ ͡°)

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4 years ago, # |
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For the problem Div1 B, Can anybody give a proper proof for the third condition i.e. s[1]=s[n] and s has at least three different characters. Thanks in advance.
EDIT: Don't bother, I got it.

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3 years ago, # |
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weee this content has so many upvote :smiling_face_with_3_hearts:

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