### ExplodingFreeze's blog

By ExplodingFreeze, 3 weeks ago, Thank you for participating! We hope you enjoyed the contest.

1605A - С.А. Отклонение

Authored and prepared by JeevanJyot

Hint 1
Hint 2
Hint 3
Solution
Solution [c++] (JeevanJyot)
Solution [Kotlin] (ExplodingFreeze)
Solution [Python] (AshishGup)

1605B - Реверс сорт

Authored by Ashishgup and prepared by JeevanJyot.

Hint 1
Hint 2
Solution
Solution [c++] (JeevanJyot)
Solution [Kotlin] (ExplodingFreeze)
Solution [Python] (AshishGup)

1605C - Доминантный характер

Authored by Ashishgup and prepared by JeevanJyot.

Hint 1
Hint 2
Hint 3
Solution
Solution [c++] (AshishGup)
Solution [Kotlin] (ExplodingFreeze)

1605D - Деревонумерация

Authored and prepared by the_hyp0cr1t3.

Hint 1
Hint 2
Hint 3
Hint 4
Hint 5
Solution
Solution [c++] (the_hyp0cr1t3)
Solution [c++] (AshishGup)
Solution [Kotlin] (ExplodingFreeze)

1605E - Уравнитель массивов

Authored and prepared by JeevanJyot.

Hint 1
Hint 2
Hint 3
Solution
Solution [c++] (JeevanJyot)

1605F - ПалиндORм

Authored by ExplodingFreeze and antontrygubO_o and prepared by ExplodingFreeze.

Hint 1
Hint 2
Hint 3
Hint 4
Hint 5
Solution
Solution [c++] (ExplodingFreeze)
Solution [c++] (antontrygubO_o)
Solution [Kotlin] (ExplodingFreeze) Tutorial of Codeforces Round #754 (Div. 2)  Comments (73)
 » Auto comment: topic has been updated by ExplodingFreeze (previous revision, new revision, compare).
 » In order to solve problems like this problem C, can anyone suggest me anything ? I have solved a lot of 'C' of div2(from another acc 'The_mysterio') , doesn't feel like it helped in this problem. It is observation I understand but may be something else in also needed????
•  » » another approach for C 135143211
•  » » Just wondering, is there a reason behind having an alt account just for solving problems? Like you could already do that with your main account.
•  » » » Two accounts just to hide my pain . Any advice for developing the hindsight for such 'C' s?
•  » » 3 weeks ago, # ^ | ← Rev. 7 →   .
•  » » solving small random test cases and finding out an observation for the problem is a good technique.
•  » » » OK.
 » Thanks for giving hints before the exact solution . It improves the thinking process .
 » For me, D was a really great problem.I was able to find out Starting at any node should guarantee a win for her. And I can solve this by implementing bipartite concept. But could not implement in time :(And yay, I won't be able to participate in Div 3 anymore. I am a blue now :)
•  » » Congrats & strong performance :)! But: You still can participate in div3 though
•  » » » Yeah, But it won't be rated for me :) Thanks.
 » 3 weeks ago, # | ← Rev. 3 →   Also my apologies for F being too tough for a Div2 round. While I expected it would be tougher than an average Div2F I didn't expect it to go unsolved during the round.Additionally, congratulations to maspy for managing to solve it (135175668) shortly after system testing concluded, before the editorial was posted.
 » Bitterly missed the 'traditional' questions today (DP, Seg-Tree based, Bin-Search, etc.) that would otherwise cushion the fall. Great contest though, ig.
•  » » You could try AtCoder Beginner Contest.
 » Thanks for the quick Editorial.
 » 7 length substring!!!!!!!!!!I will kill ya!!
 » 3 weeks ago, # | ← Rev. 2 →   i like editorial with hintsbut not this one for E<> or << solve for one query>> i think if i could solve for one query, i could solve the main problemjoking? do you think this helps? really fun man
•  » » 3 weeks ago, # ^ | ← Rev. 3 →   Just anecdotal evidence: I solved this task in the contest, and my first attempt solved the problem for 1 query (giving me $O(n \cdot q)$ complexity, which is not enough). Having implemented that, I was able to find the the optimisation needed afterwards. So I'd say, that one is a valid hint. :) I guess the problem with hints for E is, that it is a pretty technical task. Like, you use the linearity between input and output to do some linear algebra magic and then you have to work with cases with absolute values. Hard to pinpoint the thing to a "central idea".Edit: I think a two very good tips would've been: Solve this by going from left to right and changing each $a_i$ to $b_i$. Notice, later operations on $i$ don't change numbers on positions smaller $i$.Look at $a=(1,0,0,...)$ and $b=(0,0,0,...)$ and for each position $i$ count which operations were perfomed. Notice, that at each position either the first operation is performed exactly once, or the second operation is performed exactly once or no operation is performed.
 » In Problem A you can also get the same result, by noticing, that the sum $S=a_1+a_2+a_3$ is invariant under the operations. With $a_1+a_3=S-a_2$ we obtain $d(a_1,a_2,a_3)=a_1+a_3-2\cdot a_2=S-3\cdot a_2$. So it's enough to check $d(a_1,a_2,a_3)\mod3=S\mod3$.
 » Thank you for great contest! B and C were amazing :)
 » really helpful hints for E :|
 » C:What is the wrong case? https://codeforces.cc/contest/1605/submission/135143742
•  » » In this test your code gives 2 expect 31 7 abbbaba 
•  » » Try 1 2 ba Should give $-1$, gives $2$. Your code "finds" and accepts the substring a as a valid string for check(2)
 » I lost 50 rating, but I thought the contest was really cool, and I'm really happy about the hint-based editorial :)
 » Super nice and enjoyable problems, I really liked problem D. The cool thing about this problem was how some observations could lead to a neat and easy to code solution, I really loved it.hope to see more similar problems in the future. And thanks a lot for the round.
 » In case you guys prefer video solutions, here are the solutions to the first 4 problems: Solutions
 » can we do problem C using sliding window ?How?
•  » » for reference: https://codeforces.cc/contest/1605/submission/135226082
»

# include

using namespace std; int min(int a,int b){if(a>b)return b; else return a;} void swap(long long int &a,long long int &b) { int temp=a; a=b; b=temp; } int main() { int t; cin>>t;

while(t--)
{
int n;
cin>>n;
string s;
cin>>s;

int m=2e8;

int i=0,j=0,a=0,b=0,c=0;
for(int i=0;i<n;i++)
{if(s[i]=='a' &&  s[i+1]=='a')
{m=2;
break;}
}

while(j<n)
{
if(s[j]=='a')
{a++;
j++;}
else if(s[j]=='b')
{b++;
j++;}
else
{c++;
j++;}

if(j-i+1>2)
{
if(a>b && a>c)
{
m=min(m,(j-i));
}
else
{  while(s[i]!='a')
{
if(s[i]=='b')
{b--;}
else
{c--;}

i++;
}
if(i>j)
j=i;

}

}

}
if(m==2e8)
cout<<-1<<endl;
else
cout<<m<<endl;

}

return 0;

}

why am i getting RUNTIME ERROR and on which testcase it is giving WA?

 » In problem C: I know "the string must look like "a??a??a??a??a" " is correct instinctively. But I could not prove it. Can anyone give me a provement of why the smallest substring should never contain something like "a _ _ _ a"?
•  » » Because you have 3 spaces between the 2 a'sIf the 3 spaces does not contain any more $a$, then "one" of the following 2 conditions are always true - $count(a) = count(b)$ $or$ $count(a) = count(c)$And we don't want that.And if it does contain at least one more a, then you can always form an answer with at most length=3 (And this will be a better answer)
•  » » » Thanks a lot!
•  » » First of all a___a alone is not sufficient to a be a valid string because 3 elements in middle and atleast two would be same. so you need more characters to build a valid string. Now think of something like you could not make up a valid string here, so if a valid string were to exist, then you would have to make up for the fact you left spaces in between, and add some extra a's, then again as you are having two a's at a distance of atleast 4, so adding a's in this way...then the same problem comes. So to have those extra a's you would need to have those a's with dist < 4.Hope this helps. This was my logic behind it. But proving mathematically.. :((
•  » » » Thanks! It is inspiring to me.
 » $x\oplus y\nleq\min(x,y)$ for D is a typo right? $\nleq$ should be $\leq$
•  » » Yup, that's a typo, it will be fixed shortly.
 » I think there is a typo: (Thus, if MSBx=MSBy then x⊕y≰min(x,y).), is not the xor will be always less than min(x,y). correct me if i am wrong.ExplodingFreeze
•  » » You are correct, it should be $\leq$, the typo will be fixed in a bit
 » LOLSolved A using ternary search xD
 » 3 weeks ago, # | ← Rev. 2 →   I've got a little confusion according to tutorial of problem E.$|cx + d|$ = $cx + d$ exist when $cx + d \geq 0$which is $cx \geq -d$.When $c > 0$, the range of x will be $x \geq -\frac{d}{c}$When $c < 0$, the range of x will be $x \leq -\frac{d}{c}$But in the tutorial, it says $|cx + d| = cx + d$ $x \geq -\frac{d}{c}$Could anyone explain it to me?
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   JeevanJyot I think this might be a mistake in your tutorial. I see that you change $|cx + d|$ into $|-cx - d|$ if c is negative in your code, you should write it in your tutorial.
•  » » » Apologies for that. I will correct that in a while.
 » can anyone tell me what is the meaning of this statement? if a1+a3−2⋅a2≡2 mod3, then the minimum value of d(a1,a2,a3)=|2−3|=|−1|=1 like how tj=his is coming |2−3|=|−1|=1
•  » » Well, how I thought about it isa1 + a3 — 2*a2 = know if we add 1 to either a1 or a3 and subtract 1 from a2a1 + a3 + 1 — 2 * (a2 — 1) = a1 + a3 + 1 — 2 * a2 + 2 = k + 3Similarly when be subtract 1 from wither a1 or a3 and add 1 to a2a1 + a3 — 1 + 2*(a2 + 1) = a1 + a3 — 1 + 2*a2 — 2 = k — 3We can perform these operation any number of times so we can minimize our kSo, if we are given some k and we can add or subtract 3 from it any number of times.If k mod 3 = 2 Then we can simply subtract three again to make it -1 Thus (2-3) ≡-1 ≡ 2 mod 3And we are taking absolute so min( |-1|, |2| ) = 1Might not be entirely correct, but worked for me
•  » » » 3 weeks ago, # ^ | ← Rev. 2 →   ahhhh got it got it thanks for the help and for the reply!
 » Thanks for the hints,it can help a lot in solving such interesting problems."Programming is not just to write a code, it is to understand and proof your solution".
•  » » 3 weeks ago, # ^ | ← Rev. 3 →   Of course
 » C->D(1400->2100) shouldn't there be a problem of 1700 in between(considering div2)
 » Thanks for fast edutorial and giving hints before the actual solution...it has helped me in thinking more
 » Excuse me! But,I think that your solution of E has a mistake. Maybe your theory of sort is not right. Because "c" can be negative.
•  » » You can change c into positive because $|cx + d|$ = $|-cx - d|$. Although he didn't write it in the tutorial, you can see it in his code
•  » » » Oh,I get it. Thanks a lot!
•  » » » 3 weeks ago, # ^ | ← Rev. 2 →   Hello,wxy2005 I did not understood the concept of the problem E last lines of editorial This opening of |cx+d| thing .Like why we took into consideration the value of c because as it is a variable it can be positive or negative. Why the sign of c matters?Can you explain it fully ?
•  » » » » 3 weeks ago, # ^ | ← Rev. 2 →   The sign of c does not matters. The only thing that matters is the sign of $(cx + d)$ because what we want to know is $\sum |c_ix + d_i|$.We know that $|c_ix + d_i| = c_ix + d_i$ when $c_ix + d_i \geq 0$ are hold, and $|c_ix + d_i| = -c_ix -d_i$ when $c_ix + d_i < 0$ are hold.If $c_i$ is positive, the inequality $c_ix + d_i \geq 0$ holds when $x \geq -\frac{d_i}{c_i}$, but if $c_i$ is negative, it will holds when $x \leq -\frac{d_i}{c_i}$To avoid this, we found that $|c_ix + d_i| == |-c_ix - d_i|$, so we can change $|c_ix + d_i|$ into $|-c_ix - d_i|$ if $c_i$ is negative.After doing this, we can sort all $c_i, d_i$ in the order of $\frac{d_i}{c_i}$. After doing this, we can used binary search to find a position that for all $c_i, d_i$ before it $c_ix + d_i \geq 0$ and for all $c_i, d_i$ after it $c_ix + d_i < 0$
•  » » » » » No, |cx+d| = opening of this modulus operator as c can be negative then you told the new condition of x above in the comments. So , basically the range of x in the editorial where binary search is applied.So what exactly range of x binary search is applied upon? wxy2005
•  » » » » » 3 weeks ago, # ^ | ← Rev. 3 →   Now, i understood.You devoted your time to my question and replied it. Heartily thanks to you bro. wxy2005
 » 3 weeks ago, # | ← Rev. 2 →   JeevanJyot can you please explain " \n"[i+1 == ans.size()]?
•  » » Basically i+1 == ans.size() will be false (i.e. $0$) for all values of $i$ except its last value. So it will print the $0$-th character of the string " \n" which is ' '.For the last $i$, the condition will be true (i.e. $1$). So it will print the $1$-st character of that string which is '\n'. In a nutshell, it's just a one-liner way of printing some space-separated integers with a '\n' in the end.
•  » » » Thanks！
 » Thanks for the amazing contest! There´s a typo in the editorial of problem E, it says "B4−2" and it should be "B4-b2".
•  » » Thanks for pointing it out. Corrected it.
 » 3 weeks ago, # | ← Rev. 2 →   Which approach is better for practicing problems in problemset: topic-wise or difficulty-wise?
 » Time Limit is too strict for C. I used a map to keep a count on the frequency of a, b and c and it resulted in TLE :(
 » 3 weeks ago, # | ← Rev. 2 →   Regarding problem C, I found 13 minimal length substring "abbabbaccacca" which satisfies the properties mentioned in the problem.But in the tutorial it is said that there can be atmost 7 length substring which can satisfy the property.
•  » » If you take characters [3, 9] of this string of length 13 you end up with “abbacca”, which is a string of length 7 satisfying the properties.
 » I solved E with complexity $O(nlogn+q+2*A_{max})$ breaking the sum of $|c*x+d|$ into the difference between $|c*x+d|-|c*(x-1)+d|$ and using two prefix arrays but I really liked the editorial approach as it doesn't depend the value of $A_{max}$. :)
 » problem c : what is wrong in my code? https://codeforces.cc/contest/1605/submission/135505427
•  » » your code fails for test 1 4 abba
 » I really like the problem F and thank you for the hints! ^ω^
 » We can assign all integers from 1 to n having the i-th bit as MSB to a white node if the i-th bit is set in w, and assign all the remaining integers to black nodes.I am not able to prove that no matter what the value of n, we can use w to assign nodes to white or black.I understand it works, but I am not able to get how it works? Any hints :)
 » Is it possible to solve problem D (Treelabeling) via centroid decomposition? Motivation on why I believe it is possible: when you divide the tree into at least 2 components and label the centroid with number whose MSB is k, then you can label the centroids of the remaining components with some numbers whose MSB is greater than k.What have I tried: let's call the depth of centroid as the depth of the recursion when we found it (for example first centroid has depth 0, centroids of the remaining components have depth 1, after that decomposition centroids of those remaining components have depth 2, and so on...). Let's keep nodes in lists according to their depth sorted by the number of nodes in their components (when they become centroid). In the k-th list, we will pick first 2^k nodes and label them with numbers with MSB k. If there are more than 2^k such nodes, the rest we shall move to tier (list) k+1. Unfortunately this doesn't work, but I can't find the counter example :(