### Stepavly's blog

By Stepavly, 7 weeks ago, translation,

1506A - Strange Table

Problem author: sodafago

Editorial
Solution

1506B - Partial Replacement

Problem author: MikeMirzayanov

Editorial
Solution

1506C - Double-ended Strings

Problem author: MikeMirzayanov

Editorial
Solution

1506D - Epic Transformation

Problem author: MikeMirzayanov

Editorial
Solution

1506E - Restoring the Permutation

Problem author: MikeMirzayanov

Editorial
Solution

1506F - Triangular Paths

Problem author: MikeMirzayanov, Stepavly, Supermagzzz

Editorial
Solution

1506G - Maximize the Remaining String

Problem author: MikeMirzayanov

Editorial
Solution

• +65

 » 7 weeks ago, # |   +22 Problem C was a direct Longest Common Substring question
•  » » 7 weeks ago, # ^ |   +276 Thanks for sharing such a valuable insight with us.
•  » » » 7 weeks ago, # ^ |   +11 Your sarcasm is very RATIST
•  » » » 7 weeks ago, # ^ | ← Rev. 2 →   +4 You have high EQ
•  » » » 7 weeks ago, # ^ |   +5 Your EQ is really high.
•  » » 7 weeks ago, # ^ |   +1 Thanks man!!
•  » » 7 weeks ago, # ^ |   0 Yes,It's for sure.
 » 7 weeks ago, # |   +4 Nice problems. Thanks for such a wonderful contest.
 » 7 weeks ago, # |   +3 very interesting tasks!
 » 7 weeks ago, # |   +12 Spoilers are broken for me.
•  » » 7 weeks ago, # ^ |   0 for me too
•  » » 7 weeks ago, # ^ |   0 It is very annoying. Please MikeMirzayanov fix.
 » 7 weeks ago, # |   0 tutorial d is not clear
•  » » 7 weeks ago, # ^ |   +5 Consider the frequency of the most frequently occurring element. Let it be $mx$.Now, we will try to pair these $mx$ elements with rest of the elements, i.e, $(n-mx)$.$if(mx\,<=\,n-mx)\,ans = 0 \newline else\,ans\,=mx\,-\,(n-mx)$Also note that if $n$ is odd then minimum possible answer is $1$$if(n\,\,is\,\,odd)\,ans=max(ans, 1)$
•  » » » 7 weeks ago, # ^ |   0 I used the same approach but I got TLE on test case 7. What is the problem? Can someone help me?
•  » » » » 7 weeks ago, # ^ |   0 Same problem here , I am not able understand why I got TLE on testcase 7 , please anybody can explain it...
•  » » » » » 5 weeks ago, # ^ |   0 unordered_map m; m.reserve(1<<10); m.max_load_factor(0.25); By adding these two lines at after declaring unordered_map, you can get rid of TLE. For explanation checkout this blog.My solution using unordered_map: http://cf.yanyanlongxia.cn/blog/entry/21853
•  » » » » » » 5 weeks ago, # ^ |   0 Thanks brother
•  » » » » » » 4 days ago, # ^ |   0 Bro thanks for this advice!
•  » » » » 7 weeks ago, # ^ |   +20 Your logic is correct. The reason that you are getting a TLE verdict is because you have used an "unordered_map". By using an unorederd_map , you are expecting an average of O(1) time complexity. However,as you might know, unordered_map works on the concept of "hashing". So,for that particular test case(T.C-7,where your code gives a TLE verdict), there must be collision between hash values in the map,giving you an actual time complexity of O(n^2) as opposed to your expected O(1) time complexity. Judging by the constraints,you may easily see that O(n^2) is not sufficient to pass the tests. You may use "map" instead,which will give you a time complexity of O(log n).It will easily pass the tests. You may refer the given link below for more understanding on the above topic: http://cf.yanyanlongxia.cn/blog/entry/62393 Also,here is a link to my solution in case you may face any problems: http://cf.yanyanlongxia.cn/contest/1506/submission/111029797
•  » » » » » 7 weeks ago, # ^ |   +1 thanks a lot bro i also did the same mistake and i had no clue why my solution was wrong
•  » » » » » 7 weeks ago, # ^ |   +1 Yeah! Thank you for your valuable reply...
•  » » » » » 7 weeks ago, # ^ |   +1 Thnak you so much :) ....even editorial is containg unordered map
•  » » » 7 weeks ago, # ^ |   0 why is ans 0 if mx <= n — mx. This basically means the max frequency is lesser than rest so it will be paired easily but after fixing these mx elements, how do i know the left elements will be distinct and i will be able to eliminate all? or 1 will be left in case of odd. can someone explain?
•  » » » » 7 weeks ago, # ^ |   0 Left element must be distinct blc you find max frequent element so not any other number have frequency more than this so you can eliminate all numbers if there are total even numbers ...you can also check it via taking some test cases so you have very clear view...
•  » » 7 weeks ago, # ^ |   0 Tutorial for D step1: first find frequency of elements using map...then put them into a priority queue of pairs.. (priority queue will contain highest freq on its top as first — its property of priority queue containg pairs)step2: within while we will pop out first two pairs of priority queue..we will decreament their first value (its like we are choosing a pair from these two combination) and if they dont become 0 we will push themhere i am taking a confusing case ex. array size 6 , array elements are arr[]={1,2,3,1,2,3} 1's freq is 2 2's freq is 2 3's freq is 2 after 1st iteration.... 3's freq is 2 1's freq is 1 2's freq is 1 after second iteration.... 2's freq is 1 3's freq is 1 1's freq is 0 after 3rd iteration ...... all will become 0
 » 7 weeks ago, # |   0 Please put the code in spoiler tag.
•  » » 7 weeks ago, # ^ |   +1 Seems like spoilers are broken now
 » 7 weeks ago, # | ← Rev. 2 →   -6 Hey, MikeMirzayanov Just wanted to bring before you plagrism of two users Both of these Users don't know whether same persons or different has submitted solution of Problem A and B with minor Changes. Please Do look at their submission. Even their template is same. Since Even submitting Solutions from alternate account is clear voilation of policy please Review their submission. Maybe they would be same person just submitting solution from one account to confirm its correctness and escape penalty, which is voilation of Rule and needed to be punished if really found guilty. Their Submissions: Problem A: 110996666 and 111008607Problem B: 111007814 and 111006918
•  » » 7 weeks ago, # ^ | ← Rev. 4 →   -40 MikeMirzayanov plag_report Yeah my bad, but I have an explanation for that which I think might be justified. Since the past 3 contests I've been facing issues with the CodeForces server being down, as might be evident from the bad results of my past 2 contests as well as the fact that I could not give the Division 2 immediately preceding today's contest. Thus, to prevent a negative effect on my rating I started today's contest with my alternate account but switched back to my original one after facing no issues for about half an hour.I had no idea that submitting from 2 accounts is also a violation of Rule, and I thought that since I could easily prove the 2 accounts to be mine, I would be able to show that the code is completely original in case someone said otherwise.I can assure you that it wasn't a case of trying to avoid a penalty because I did get penalties in my D and E as well. If I had been trying to avoid penalties by testing solutions with my alt account, then I obviously wouldn't have stopped that for the harder problemsMoreover, I submitted the solution for A from my main account 15 minutes later, which covers up the 1 penalty advantage that I would've received in problem B
•  » » » 7 weeks ago, # ^ | ← Rev. 2 →   +10 Do you even read the rules of participation before clicking on that "Register" button while registering for participation in a contest ? It's clearly written there that you can't use multiple accounts
•  » » » 7 weeks ago, # ^ |   +2 You can use second and third websites for contests. [Second website(http://m2.cf.yanyanlongxia.cn/) Third Website]
 » 7 weeks ago, # | ← Rev. 3 →   +5 my solution for D sorry for my englishlet MAX be the maximum number of times a number occurs. if MAX>n-MAX, then the answer is MAX — (n-MAX) because we can delete this number with all the others, but we can't delete it with ourselves. otherwise, the answer is n%2, because we can delete the remaining numbers up to the number of our number, and then delete them with our number. well, if n%2 == 0, then we can delete everything, otherwise it is clear that we will not be able to delete one element because we delete 2 elements each
•  » » 7 weeks ago, # ^ |   +2 how can we prove if n%2 == 0 and MAX<=n-MAX then every number will get paired up
•  » » » 7 weeks ago, # ^ |   -7 you can easily prove this by induction.
•  » » » » 7 weeks ago, # ^ |   0 How would you prove it? Please provide a proof.
•  » » » » » 7 weeks ago, # ^ |   0
•  » » » » 7 weeks ago, # ^ |   0 can u pls tell the proof??
•  » » » » » 7 weeks ago, # ^ | ← Rev. 2 →   0 For base case: Let's say we have two groups ($a$ & $b$) of elements initially. $freq[a]=size/2$ $freq[b]=size/2$ So we can pair them up easily. Now we assume that we can pair $k$ groups of elements.Now for the $k+1$ th group, No of elements in this group $\leq$ $totalSize/2$. So we can always break a pair and make two new pairs with the elements of $k+1$ th group. Although I assumed that size of $k$ groups is even, but you can see that if it was odd then we could pair one element of $k+1$ th group with that left element. You can extend this proof to $size$ being $odd$ where you could make a argument that we will be left with one element after pairings are done. And in fact this proof is for both $odd$ and $even$ size combined.
•  » » » 7 weeks ago, # ^ | ← Rev. 2 →   +41 Sort the array and pair up $a_i$ with $a_{i+n/2}$. Since every number appears no more than $n/2$ times, these two numbers are always different.
•  » » » » 7 weeks ago, # ^ |   0 Easy to understand!
•  » » » 7 weeks ago, # ^ |   0 Prove by contradiction-> Let there be some numbers left. Without loss of generality, we can assume that there are x 1's left. Let's assume initially there were y 1's present in the array. So, we can safely say that (y-x) 1's were paired with some numbers. So, there are $n-y-(y-x)$ elements left which are neither 1's nor paired to any 1. So, there were $\frac{n-y-(y-x)}{2}$ pairs involving no 1. So, we will try to break $\frac{x}{2}$ of those pairs and pair each number with the remaining $x$ 1's. Now, we just have to prove that $\frac{x}{2}\leq\frac{n-y-(y-x)}{2}$. $x\leq n-y-(y-x)$ $2*y \leq n$Now, as the most frequent elements occur less than $\frac{n}{2}$ times, so this condition is satisfied. Hence proved! Voila!
•  » » » » 6 weeks ago, # ^ |   0 Thanks
•  » » » 7 weeks ago, # ^ |   0 Since now MAX <= n-MAX, you could pair numbers in n-MAX. Whenever you delete a pair, n-MAX => n-MAX-2. Continue this process until MAX == n-MAX-2*pair_nums.[Noticed that there must a pair_nums which fits the equation. If not, then there is a number which appears (n-MAX-2*pair_nums) times and MAX < (n-MAX-2*pair_nums), however MAX is the max appear times.]Then we could delete each number in MAX with others.
 » 7 weeks ago, # |   -8 Learnt something new today — s.lower_bound(x) is MUCH MUCH faster in set than lower_bound(s.begin(),s.end(),x). Wasted an hour and will cost me my rating but definitely wont make this mistake ever again. Thanks :)
•  » » 7 weeks ago, # ^ |   0 Blog about it! Luckyli I faced this while practicing. And today that experience helped me!
 » 7 weeks ago, # |   0 Can G be solved with stack??
•  » » 7 weeks ago, # ^ |   0 yes, u can maintain the next greater element for every element and then compute the answer using prefix sums...
•  » » 7 weeks ago, # ^ |   0
 » 7 weeks ago, # |   0 Good Problems, Thanks for such a wonderful div 3 :)
»
7 weeks ago, # |
-41

# HAPPY CODING

•  » » 7 weeks ago, # ^ | ← Rev. 2 →   +50 Just a friendly advice, you should try and improve your coding skills and try streaming later when you are better at coding.
•  » » » 7 weeks ago, # ^ |   -56 just a friendly advice too : Don't judge a Book by its cover and Oil your own machine ..
•  » » » » 7 weeks ago, # ^ |   +50 Easy dude, just tried to help you not argue with you. Fine, I lose, you are one of the best coders and are really great at coding. Okay?
•  » » » » » 6 weeks ago, # ^ |   0 What a reply dude! You had me laughing!!
 » 7 weeks ago, # |   +10 I guess problems were easier than the editorial :)
 » 7 weeks ago, # |   0 Can anyone please explain the problem 1506G — Maximize the Remaining String in simpler way?
•  » » 7 weeks ago, # ^ | ← Rev. 2 →   +10 Here's my pretty basic approach:For the first letter, can we make it Z? We can if we have a Z, and every other unused letter appears after the first instance of Z. If not, try Y, X, W, etc.Then repeat this for the second letter, third letter, fourth letter, discounting the letters we have already used, and strictly considering positions after our last used letter. We finish when there are no more unused letters.Here's my submission (python). It's reasonably easy to follow.
•  » » » 7 weeks ago, # ^ |   +5 Thanks a lot for this approach! Was a good bitmask exercise for me in c++.
•  » » » 5 weeks ago, # ^ |   0 Me still no understand :'(
•  » » 7 weeks ago, # ^ | ← Rev. 2 →   0 Made a video on this, you can give it a watch. http://youtu.be/NMSiu-VNfYU
•  » » » 2 weeks ago, # ^ |   0 Thanks a lot! Really helped!
 » 7 weeks ago, # |   0 Stepavly for problem E, i think u forgot to mention approach for lexicographically maximal string. I cant find it "If we want to build a minimal lexicographic permutation, we need to build it from left to right by adding the smallest possible element. If q[i]=q[i−1], so the new number must not be greater than all the previous ones, and if q[i]>q[i−1], then necessarily a[i]=q[i]. q[i]q[i−1], then a[i]=q[i], otherwise we put the minimum character that has not yet occurred in the permutation."
•  » » 7 weeks ago, # ^ |   0 To get the maximum you can do something very similar.If q[i] > q[i-1], again we have q[i] = a[i]. When this happens, add all integers between q[i-1] and q[i] to a maximum priority queue. For elements where q[i] = q[i-1], a[i] is the element at the top of the maximum priority queue.
 » 7 weeks ago, # |   +3 If c1=c2, then if (r1+c1) is even, then the cost is r2−r1, otherwise the cost is 0;Shouldn't that be $r1 - c1 = r2 - c2$ instead of $c1 = c2$?
 » 7 weeks ago, # |   +9 Can someone plz give a more beginner friendly explanation to problem G. I cant understand the one given in the editorial.
•  » » 7 weeks ago, # ^ |   +6 Hope this helps
•  » » 7 weeks ago, # ^ |   +9 Maybe 111074804 this can help.
•  » » 7 weeks ago, # ^ |   0 stack if(s[i]>stack.top()) { then check is their any element in string s which equal to stack top ans pos also >i then perform stack pop operation } 
 » 7 weeks ago, # |   0 For D, it is obvious and correct that we need to subtract from the maximal (largest). But does taking off the second largest matter? I thought it did but I came up with some examples and it seems like the second item can be any number from the list. It seems like we don't need to use the second largest. Can someone provide a counter example for this?
 » 7 weeks ago, # |   0 boooooooooooooooooring round
 » 7 weeks ago, # | ← Rev. 2 →   +2 what will be the time complexity of editorial of problem d? since the value of ai can go up to 1e9;update: sorry I just forget, we are dealing with frequency. :( How can I be so dumb!
•  » » 7 weeks ago, # ^ |   +1 We are not dealing with ai. We are dealing with the frequency array of a. Since total number of elements <=2e5,the sum of elements of freq array doesn't exceed 2e5.
 » 7 weeks ago, # |   0 Can someone help me with C.. My code
 » 7 weeks ago, # |   +46 When Problemsetters don't want the police to bother them! Credits — Problem G(For those who don't know, "AEZAKMI" is a cheat code in GTA San Andreas to escape from the police permanently)
•  » » 7 weeks ago, # ^ |   0 Glad someone pointed that.
 » 7 weeks ago, # |   +4 Problem G is the same as http://leetcode.com/problems/remove-duplicate-letters/
 » 7 weeks ago, # |   0 I had the exact idea for D. But I didn't code it since I thought it would give TLE. Can someone explain why the solution for D won't TLE?
•  » » 7 weeks ago, # ^ |   0 We can see at max we have to do n removal operations and n insertions operations in the priority queue. We dont want to deal with the numbers exactly, instead with their frequency. Insertion and Removal operations take O(n logn) time where n is the number of elements already in the priority queue.Also it isn't necessary to use priority queue. I used multiset to solve the same problem but the idea was similar. Have a look. 111083842.
•  » » » 7 weeks ago, # ^ |   0 okay i get this point but how could you be sure that deleting the top two elements from the multiset would fail and the other way of deleting one from the top two values in multiset would work ??1 6 2 3 2 1 3 1 surely, our code with former logic would fail on this testcase, but pass the rest of them. What if this testcase wouldn't be there, then how would you reach to the above conclusion that the latter approach of decrementing the top two values by one would only work, and the former approach of deleting the second largest value from the top value at one go would give WA
•  » » 7 weeks ago, # ^ |   0 I thought of the same thing during the contest. But the sum of n over all test cases wont exceed 2e5. So it passes.
•  » » » 7 weeks ago, # ^ |   0 Oh yes, sum of n never exceeds 2*10^5. So, how do we write the time complexity of the whole solution? Definitely not O(t * n).
•  » » 7 weeks ago, # ^ |   0 i coded it and got TLE, XD. Here in editorial they used 2 elements for priority_queue but i used only one. Though priority_queue does not work on hashing still i don't know why it gave me TLE //This is my code. int n; cin>>n; vector a(n); unordered_map cnt; for(int i=0;i>a[i],cnt[a[i]]++; priority_queue pq; for(auto u:cnt) pq.push(u.second); while(pq.size()>1) { int hi=pq.top(); pq.pop(); int lo=pq.top(); pq.pop(); if(hi>1) pq.push(hi-1); if(lo>1) pq.push(lo-1); } int ans=0; while(pq.size()) { ans+=pq.top(); pq.pop(); } cout<> q; //changes are here only int n; cin >> n; map v; for (int i = 0; i < n; i++) { int x; cin >> x; v[x]++; } for (auto u: v) { q.push({u.second, u.first}); } while (q.size() >1) { auto hi = q.top(); q.pop(); auto lo = q.top(); q.pop(); hi[0]--; lo[0]--; if (hi[0]) { q.push(hi); } if (lo[0]) { q.push(lo); } } int ans = 0; while(q.size()) { ans+=q.top()[0]; q.pop(); } cout << ans << "\n"; 
•  » » » 7 weeks ago, # ^ |   0 it was the same in my case. when i used unordered_map, it gave me TLE. changing it to map gave AC. idk why it happens. is the inbuilt hashing function that bad?
•  » » » » 7 weeks ago, # ^ |   0 //i think to avoid that problem need to use below custom_hash(originally from neal's blog) struct custom_hash { //ex define like below //unordered_map safe_map; static uint64_t splitmix64(uint64_t x) { // http://xorshift.di.unimi.it/splitmix64.c x += 0x9e3779b97f4a7c15; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9; x = (x ^ (x >> 27)) * 0x94d049bb133111eb; return x ^ (x >> 31); } size_t operator()(uint64_t x) const { static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count(); return splitmix64(x + FIXED_RANDOM); } }; 
 » 7 weeks ago, # |   +1 Weak pretests for problem E. Well, it was ultimately my mistake as I ignored the time complexity of my code
•  » » 7 weeks ago, # ^ |   0 Yeah, felt so bad to get TLE 12 hours after it was initially accepted. I didn't expect my solution to pass but it still hurt.
•  » » » 7 weeks ago, # ^ |   0 Same here, I slipped from 129 to 628 in standings. yet it was a nice learning
•  » » 7 weeks ago, # ^ |   0 Me too I though the complexity of my code was o(n) but on closer examination on some cases its o(n^2). Many people had submitted such a solution and had their solution hacked.
 » 7 weeks ago, # |   0 For question D,why can map pass this questionAccepted but use unorderd_map will TLE
•  » » 7 weeks ago, # ^ |   0 Have a read. Blog !
•  » » » 7 weeks ago, # ^ |   0 thanks for the link of the blog, never trust unordered_map!!
•  » » » 7 weeks ago, # ^ |   0 Thanks
•  » » 7 weeks ago, # ^ |   0 using unordered_map i got ac when contest is running. After the contest its showing tle on a test case
•  » » 7 weeks ago, # ^ |   0 Ordered map aka usual "map" is sorted as a default so any operation performed on it is O(log(n)) whereas unorderedmap will cost O(n) for each operation
 » 7 weeks ago, # |   0 Can anyone help me with E. Why my code got TLE? Code
•  » » 7 weeks ago, # ^ |   0 while(used[r]) r++;this part of your code is the reason for TLE which makes the overall time complexity of your code as O(N²)
•  » » » 7 weeks ago, # ^ |   0 can you explain why there are collisions on using unordered_map in problem [D].If my hash-map is just used for storing frequency of every unique number in the array, then how can there be any collisions in keys ?????
 » 7 weeks ago, # |   0 My solution to Problem G has the time complexity of O(N^2), yet it has got accepted. Can anyone tell whether the testcases are weak or I have wrongly calculated my time complexity. Link to my code
 » 7 weeks ago, # | ← Rev. 2 →   0 i don't why the E problem tutorial solution seems bit lengthy to me my solution 111102730
 » 7 weeks ago, # | ← Rev. 3 →   0 Problem D Link Can anyone tell what gone wrong with my solution it giving tle now (same solution as given in editorial)But now it give tle on 7 th test case . Thanks
•  » » 7 weeks ago, # ^ |   0 use map
 » 7 weeks ago, # |   +1 Can someone explain the solution of G?
•  » » 7 weeks ago, # ^ | ← Rev. 2 →   0 suppose you start from index 0 and move towards right. It's all good until and unless you encounter a character c at position pos, such that it is not present further on right hand-side. So you have to keep c either from this location or from left hand — side (from its possible previous occurrences). What you can do is take the maximum character in sub-array s[0..pos]. Lets max is at position maxpos. Now, you have thrown s[0...maxpos-1]. Since you had to make the next character as maximum(for lexicographically max string) as possible without violating the rules. You have done right. Now start from maxpos + 1 instead of 0 and again do the same as done above and do not consider the letters previously selected. Time complexity would be O(26 * N) as for every character we are traversing one time full array in worst case.
 » 7 weeks ago, # |   0 This is my submission for D: AC Submission for D My idea is similar to that of the editorial, only difference is, in each step, I am deleting exactly one occurence of the most frequent number and exactly one occurence of the least frequent number. Its still getting ACed, Why is this correct ?
 » 7 weeks ago, # | ← Rev. 2 →   0 Why did my code for E gave TLE. As far as I know insert, lower_bound and upper_bound all are O(log n), so the overall complexity should be O(nlogn) right ??
•  » » 7 weeks ago, # ^ |   0 Hopefully this will clear your doubts — http://cf.yanyanlongxia.cn/blog/entry/76206?#comment-606743
•  » » » 7 weeks ago, # ^ |   0 Thanks, got it. Also could you tell which part of the code is taking time in this vector implementation.
•  » » » » 7 weeks ago, # ^ | ← Rev. 4 →   0 Erasing an element from a vector takes O(n) time complexity. So, the overall time complexity of your function is O($n^2$). I would suggest you clear your fundamentals especially related to the STL library and its function's time and space complexity so that you don't make further conceptual errors in implementation at least. Please refer to this — http://stackoverflow.com/questions/28266382/time-complexity-of-removing-items-in-vectors-and-deque#:~:text=Erasing%20an%20element%20in%20a,complexity%20is%20O(n).
•  » » » 4 days ago, # ^ |   0 Thanks. I didn't knew the difference between the time complexity of s.lower_bound(x) and lower_bound(s.begin(),s.end(),x). Learnt something new!
 » 7 weeks ago, # |   +13 B, C solutions with regexes instead of usual loops. Perl: B — 111112920, C — 111113262
 » 7 weeks ago, # |   0 Can anyone help what is wrong with this code ?111115251
 » 7 weeks ago, # |   0 Why this solution (in Python) for question E gives TLE, but this (in CPP) does not. Can anybody explain?
 » 7 weeks ago, # |   0 Did problem D have anti-hash testcases? I feel that because my unordered_map solution got TLE after the contest but my map solution got AC after I submitted it now.
•  » » 7 weeks ago, # ^ |   0 Same thing happened to me as well. Well it because In Unordered_map if collision increases, resulting time-complexity rises up to O(N^2). That's the exact reason for TLE for big cases. Read more about this here : http://cf.yanyanlongxia.cn/blog/entry/62393Try to use safe-map to get around this limitation of unordered_map!
•  » » » 7 weeks ago, # ^ | ← Rev. 2 →   0 Funnily enough, I know this trick. But I was kinda running out of time (had wasted a good lot of time on B), so I didn't get time to copy the snippet because I thought anti-hash cases are rarely used, and also it was very unlikely for anti-hash cases to be incorporated in a div 3 contest.Turns out they did incorporate anti-hash testcases here. So yeah, I'll keep the snippet in my main template from now on. "Learned the hard way out", yes.My problem verdicts' table looks like: A — AC B — AC C — AC D — TLE E — AC feelsbadman
 » 7 weeks ago, # |   0 Used the exact implementation as given in the editorial for Question D. But resulted in TLE as I have used Unordered Map instead of Map. Learned the hard way out.
 » 7 weeks ago, # |   0 Hey does anybody knows the scoring criteria for hacking someone's solution? I made around 15 successful hacks but my rank does not seem to have improved much.
 » 7 weeks ago, # | ← Rev. 2 →   0 Can anyone show why my E problem code got RTE? http://cf.yanyanlongxia.cn/contest/1506/submission/111127527Thanks alot
•  » » 7 weeks ago, # ^ | ← Rev. 2 →   0 Your code uses erased iterator. auto it = s.lower_bound(a[i]); if (it == s.begin()) { cout << *it << " "; continue; } -- it; if (it != s.end()) s.erase(it); res[i] = *it; // uses a deleted iterator Just modify slightly will AC(111132839). if (it != s.end()) res[i] = *it, s.erase(it); 
•  » » » 7 weeks ago, # ^ |   0 Got it, thanks alot
 » 7 weeks ago, # |   0 Can someone tell me why using unordered_map instead of map in problem D gives TLE on test case 8.TLE when using unordered_map: http://cf.yanyanlongxia.cn/contest/1506/submission/111132652Accepted when used map instead: http://cf.yanyanlongxia.cn/contest/1506/submission/111132742
•  » » 7 weeks ago, # ^ |   0 Use custom hash with unordered maps
 » 7 weeks ago, # |   0 can someone explain G again ?
•  » » 7 weeks ago, # ^ | ← Rev. 2 →   0 Just keeping adding the highest character to your string such that following conditions are met: (Suppose the index of your highest character is i in the orignal string(s). answer string is t) then i >index of last character in t also take lowest such i 3.check if s[i+1....s.size()] contains all remaining characters which are not added to your answer string t.if any of these conditions are not met move to the next lower character. continue this process till you have added all unique characters to string t
 » 7 weeks ago, # | ← Rev. 2 →   0 .
 » 7 weeks ago, # |   0 Can someone explain me this line in G :- unordered_set used(s.begin(), s.end());
•  » » 7 weeks ago, # ^ |   0 how does this make a set with members as letters of string and arrange them backwards ?
•  » » » 5 weeks ago, # ^ |   0 unordered_set -> only stores unique elements using (s.begin(),s.end()); is just saying store all the unique characters in from the start of s to the end of s in used.
 » 7 weeks ago, # |   0 So basically in problem G . we are checking if a char can be the first char finding if all unique elements left in string s can be placed after that char or not , if yes then it becomes maxC and we do that for every char in s to find the maxC that can do that which makes the string twe are also creating a filter string which has all the elements left to be placed after maxC is placed at the front , so removal of all elements before maxC and inclusion of all elements after maxC
 » 7 weeks ago, # |   0 Problem E : http://www.youtube.com/watch?v=uVfjprAFIIo
 » 7 weeks ago, # |   0 I did problem D with a very simple 2 pointer solution. Initialse left pointer,lft=0 & right pointer, rgt= ceil(n/2). Now, till you the rgt is not at the end of array, if arr[lft]==arr[rgt]: pair_count++,lft++,rgt++; else rgt++; final answer would be n-pair_count.
 » 7 weeks ago, # |   0 **** Problem E ******I wrote O(nlogn) solution for problem E using sets but it is giving TLE in 3rd test case. Here is my solution: here.Please anyone help me to find the problem in it. I am not able to figure it our.
•  » » 7 weeks ago, # ^ |   0 You wrote erase and find for a set in the loop so it's more than $O(n\ log\ n)$ .
•  » » » 7 weeks ago, # ^ |   +1 I have used only erase at first and it gave TLE so, I tried by using find and erase and again it is giving TLE.Later I have found it is giving TLE bcoz of using a normal lowerbound function, I replaced it with set lower_bound then it got accepted.
•  » » » » 7 weeks ago, # ^ |   0 Oh I see.
 » 7 weeks ago, # |   0 Hi there, I am trying to upsolve problems from this contest and I am struggling to see why I am getting the Runtime Error on test 2 for problem E. Here is the code:http://cf.yanyanlongxia.cn/contest/1506/submission/111272555Sadly I cannot see all the tests and have no clue where the error might be. Any help would be very much appreciated.
 » 7 weeks ago, # |   0 Please can someone explain statement of problem F ? Actually, I don't really understand how it is possible to pass through all N points as the edges are directed Thanks by advance for your help!
•  » » 7 weeks ago, # ^ |   0 It is guaranteed by the input of the problem, so they will always give n points which you can pass by them after sorting them by layer
 » 7 weeks ago, # | ← Rev. 2 →   0 can someone tell me where i went wrong in D -im getting tle for test case 8my submission
 » 7 weeks ago, # | ← Rev. 2 →   0 Since , Problem E is tagged with DSU any idea , how to solve it with dsu ?
 » 6 weeks ago, # |   0 Problem:1506G - Maximize the Remaining String Submission: 111543776I just used the brute for to find the maximum alphabet until a alphabet with single occurrence or all the occurrences of a alphabets occursFor ex: 'codeforces' in this in first loop I scan upto 'd' as it is an alphabet with single occurrence and then continue further. 'abbac' in this for first loop I scan upto 2nd 'b' as this is the last occurrence of 'b'. Can anyone help what is the error
 » 6 weeks ago, # |   0 [Problem E] I invented a very clean implementation of the solution to problem E. Both for max and min, maintain a pool of unused numbers; If V[i] == V[i-1], pick the most relevant element from the pool and add to solution; If V[i] != V[i-1], update the pool by all numbers in range [V[i-1] + 1, V[i] - 1], and extend both solutions by V[i]. The key point here is that the pools can be created using queue for min (we will always pick smallest non-used this way) and for max by using stack (now we will always pick the largest not used number)! This way we implement the solution in O(n) and in around 25 lines of code, max-part of it being an almost direct copy of the min-part. My solution: submission 111566779 (mal and ros are respective pools, and omal and oros are vectors for the solution to the problems).
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6 weeks ago, # |
0

I am new to code forces and I have been writing this contest in practice mode I need help! I have trouble regarding problem B this my code I didn't use the greedy or DP approach it has passed almost all tests but it was failing on some tests I have found out through the error that it might be the input where all characters in the string are '*'.but I couldn't fix this I still believe this the correct code can anybody help me through this

# include<bits/stdc++.h>

using namespace std; int main() { int t; cin>>t; for(int i=0;i<t;i++) { string s; int n,k; cin>>n>>k; cin>>s; int a[n]; for(int j=0;j<n;j++) { a[j]=0; } int l=0; for(int k=0;k<n;k++) { if(s[k]=='*') { a[l]=k; l=l+1; } } int c,d,e; c=a[0]; for (int p=0;p<n;p++) { if(a[n-1]!=0) {d=a[n-1]; break; } else if(a[p+1]==0) { d=a[p]; break; } else {d=c}

}

if(d==c||n==1) { e=1; } else { e=(d-c-1)/k + 2; } cout<<e<<endl; } return 0; }

I hope I will find a solution thanks

•  » » 6 weeks ago, # ^ |   0 The logic of your program is not correct. You cannot get correct answer with dividing by k the quantity (or positions) of asterisks. Asterisks may be located consecutively in one part and in (k-2) places in another part. Initialize array a with -1 instead of 0. Analize next asterisk position and remember position of the last asterisk you took to result.
 » 6 weeks ago, # |   0 Why for test case 7 5 **.... in problem B the answer is 3?
 » 5 weeks ago, # |   0 whats the time complexity for problem g (the code presented in this editorial
 » 5 weeks ago, # |   0 PLS HELP T-TI keep getting anything after testcase #302 wrong and have no idea why. I read the editorial and understood it, but I can't figure out whats wrong with my code since it's pretty different from the editorial solution. #include using namespace std; typedef long long ll; typedef string str; typedef char ch; /* Convert first and last * to X Go maximum distance each time to replace each * */ int tc, n, k, aP, res; str s; void solve(){ res = 0; cin >> n >> k >> s; int fs; bool first = true, found; for(int i=0;i0;j--){ if(s[l+j] == '*'){ s[l+j] = 'X'; res++; break; } } } } l++; } cout << res << endl; } int main(){ cin >> tc; for(int i=1;i<=tc;i++){ solve(); } return 0; } `
 » 3 weeks ago, # |   0 http://cf.yanyanlongxia.cn/contest/1506/submission/113624690Can someone help me in this? getting runtime on test 2 in E.