Hello Guys I have divided problems into posts according to difficulty. Hope u all don't mind this. ;) This is the part 1 of two posts of unofficial editorials for November Long Challenge. For editorials of problems CSUBQ and SEGPROD, click here. This post has editorials for problems VILTRIBE, CLRL, PERPALIN and CHEFHPAL (so many in single post :) ). Problem VILTRIBEProblem Difficulty : CakewalkProblem ExplanationGiven a string consisting of characters 'A', 'B' and '.', Output number of characters controlled by A and B. Character C controls ith character if either of following condition is true.
SolutionJust do as they asked in the problem. :) Maintain a character C (initialised to any other character of your choice other than 'A', 'B' and '.') which denote the previous controlling character encountered. Here, we are going to count characters satisfying above of the two conditions separately Beacuse if we choose to count them together, there's a risk of counting characters satisfying first condition twice, which is certainly not what we want. For example of this, try input A.A.A...B.B.B Maintain counter variables countA, countB, A, B and last where
After all this, I guess my solution is straight forward to understand. In case you feel i should clarify any doubt, Just drop a comment. :) Here's a link to my Code Problem CLRLProblem difficulty:SimpleProblem ExplanationGiven an array of numbers, we are supposed to verify whether the array is valid based on given condition. SolutionThe most simple way to think about this as following: Here, Ri denote ith element of given array. The given array is valid if and only if: For every pair of consecutive numbers R(i1) and Ri: If R(i1) < Ri : R(i1) < all elements after ith index. (type 1) If R(i1) > Ri : R(i1) > all elements after ith index. (type 2) That's it. we just need to implement it. I have used two boolean to check if any pair of type 1 and type 2 is found. Basic case : N <= 2 Answer is "YES" No matter what the elements are. (Be sure to input them or you'll get WA) For explanation, i set reward if anyone finds such a case with N <= 2 where answer is NO, He or she'll be appreciated in my next post. Good Luck :) Use max and min variable to keep record of upper and lower bound. loop from start+1 to end compare elements (i1) and ith position. if (i1)th < ith if(type1 is not found yet) found1 = true. min = ith element if type 2 is found && ith element > max, valid = false min = Math.min(min, prev) Other case works the same way. Just swap type 2 with type 2, min and max and invert the inequalities. if valid print "YES" else "NO" Link to my code Problem PERPALINProblem Difficulty : SimpleProblem ExplanationGiven two integers N and P, construct a palindrome string of length N in which every ith character is same as (i+P)th character using only characters 'a' and 'b'. If not possible, print impossible. Further, String should not contain all a or all b. SolutionFirst thing to observe is that in case P == 1 OR P == 2, ans is impossible. Reason if P == 1, valid strings are only aaaaa... (upto N times) or bbbbb... (upto N) times. But as Chef dislikes these strings, ans is impossible. if P == 2, the only valid strings can be abababab or babababa. Since N is always divisible by P, N is not odd in this case, and when N is Even, first and last character of string is never same. So we cannot get a palindrome of period length 2. ans is impossible. In all other cases, Answer always exist. Here, there exists many solutions for this problem. But I'm gonna tell which approach i used. Small string of length P will be repeated N/P times to generate the required string of length N. I generated small string as: In case P is odd, just make string ababababa till length P If P is even: (P+2)/4 times 'a' + (P/4)*2 times 'b' + (P+2)/4 times 'a' (Integer division) You may ask why i chose such a complex way to generate this string for even P. The reason is, My Choice. :D It always make palindrome string for even length >= 4. First few strings for even P would be abba, aabbaa, aabbbbaa, aaabbbbaaa Just see first half of strings => ab, aab, aabb, aaabb. My pattern make strings like that, increasing a and b one by one and then reverse it and append it to original one. Hope i didn't confused you with my small rant. Feel free to ask. Here's a link to my Code Problem CHEFHPALProblem Difficulty : EasyProblem ExplanationGiven two integers N and A, construct a string of length N using only first A characters, minimizing length of longest palindrome substring. 2<=A<=26 SolutionFirst thing to observe here is that for A > 2, One of the correct answer is "1 abcabcabc..." upto N characters because this string cannot have any palindrome of length greater than one. So that just leaves A == 2, The Special Case Here, I am making a statement, Do tell me any counter example and you will be appreciated. For N > 8, there is always a string whose maximum palindrome substring length is exactly 4. I have tested many strings using a tester program, generating strings using two characters 'a' and 'b', and found this. An alternative solution is most welcome. (If you want the tester program i used, I'd recommend you to make it yourself, it would boast your skills. :) ) The reason that this works is that for a palindrome of length >= 4, we need two pair of character matching (first with last, second with second last), so we are creating a single mismatch every time. We can't avoid palindrome length 4 after N >= 8 because consider a string aaababbb, now if we append 'a', palindrome length is 5, if you append 'b', palindrome length becomes 4. See string babbaababbaa. This string always create exactly one mismatch for substring of length > 4. (You can always find such a string with hit and trial). So, I simply stored results for N <= 8 in an array. So, i just check A>2, if yes, print abcabcabc... upto n characters. else if N <= 8, print answer precalculated (Or calculate using program if you can) else append above string repeatedly to output string till output length is exactly N. Remove extra characters in case length > N. Here's a link to my Code. As always, i wholeheartedly invite your suggestions and thank for your response to my previous editorials. asked 13 Nov '17, 19:38
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I did it in complete lazy way just find out the pattern for the length of string 11,12,13,14,15 and 16 then i continued it then i noticed there was repetition of some pattern i.e., for 11 to 15,17 to 21,23 to 27 and so on add "bbaa" next 12 to 16,18 to 22 and so on add "baab" next 13 to 17,19 to 23 and so on add "aaba" next 14 to 18 ,20 to 24 and so on add "abab" similarly for 15 to 19 add "babb" and last 16 to 20 and so on add "abba".you will get always ans 4 for string having only a & b for length having greater than 8. Here is my solution link text if you like please upvote me. answered 14 Nov '17, 22:14
Nice approach mate. I said there are many approaches, but saw first submission, that didn't use a string precomputed.
(15 Nov '17, 00:35)

In 4th Ques: For A=2 & N>8 we can also repeat the string "aababb" and mod out extra characters to get a string of exactly N length with maximum palindrome substring length = 4 My Code answered 13 Nov '17, 19:48
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Thanks for sharing... I had said, there are many strings that may work. I found the one mentioned above. :)
(13 Nov '17, 19:50)
aabbab also does the job. Link.
(13 Nov '17, 20:02)
There are total of 810 such patterns which do it. Though half are asymmetric reflections of each other.
(13 Nov '17, 20:16)

for the CHEFHPAL problem I tried writing the possible palindrome of length 16 and found a pattern A short Pythonic code answered 13 Nov '17, 21:05

For CHEFHPAL, the proof of correctness is by exhaustion. For any value of For The threeletter palindromes are For answered 14 Nov '17, 07:53

answered 14 Nov '17, 10:25

@taran_1407 Solution to problem PERPALIN can be simplified: Create array[p] and a[0]=a[p1]='a' Then fill the rest of the elements with 'b' And finally print the array (n/p) times. This saves some 'increase' hassle, probably a bit of memory too. answered 16 Nov '17, 00:31

I wasn't able to solve CHEFHPAL and by the way it's a nice editorial bro and your approach is also simple and easy to understand answered 13 Nov '17, 19:51
Glad u found them helpful. :)
(13 Nov '17, 19:54)
I have a doubt @taran_1407 I use the codeblocks ide whenever i use this code snippet ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL); Output is printed after I Give all the Inputs(if there are 4 testcases after i give all tc inputs o/p is printed) But when i don't use it Output is Printed after i give the each t.c i/p Can you tell me why this happens ??
(13 Nov '17, 20:04)
I'm sorry i can't. I don't code often in c++. Maybe @vijju123 may be able to help. :)
(13 Nov '17, 20:07)
I think you should google this out. That will be better than anything I can explain :D
(13 Nov '17, 20:13)
Ok Bro and keep up your Good work
(13 Nov '17, 20:14)
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@sagijagadish when you write cin.tie(NULL) at the beginning of the code, compiler doesn't clear the output buffer every time it prints something. Whatever you try to print is stored in the output buffer and printed only once, when the execution of code is finished. Bonus : Try using endl instead of '\n' in your code. 'endl' clears the buffer and add new line at the end.
(13 Nov '17, 20:25)
Thanks Mate. @sagijagadish
(13 Nov '17, 20:59)
Bonus: endl slows down execution a bit, try to use \n if possible xD
(14 Nov '17, 00:09)
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same approach @taran_1407 here's the code in c++ https://www.codechef.com/viewsolution/16205791 answered 13 Nov '17, 21:17
Sorry, but I'll see solutions later mate. Writing second part (even longer than the first. Though i hope anyone else on forum might be able to help you even before me.
(13 Nov '17, 21:22)

PS:2nd part will be posted soon, maybe within an hour Sorry for delay. Second part May also include POLY as delay gift. :)
Like everytime , Nice work!
Thanks Mate!
Waiting for second part and delay gift:)
Polygon pl0x <3
Not sure. I haven't solved the problem myself..
Asking someone to explain it to me. (while giving credit)
Though i would try my best to convince that person.
@droy0528, second part posted. There's a bit of delay even in delay gift :)
For PERPALIN if(N % P != 0) impossible...
It is given in problem, that N%P = 0