### Golovanov399's blog

By Golovanov399, 3 years ago,

Sorry for the issues with a couple of problems

Problem A of tc/div2 (Finding Sasuke)
Problem B of tc/div2 (A New Technique)
Problem C of tc/C div2/A div1 (Perform Easily)
Problem D of tc/D div2/B div1 (Shurikens)
Problem E of tc/E div2/C div1 (Solo mid Oracle)
Problem F of tc/D div1 (Roads and Ramen)
Problem E of div1 (A Convex Game)
• +146

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 » 3 years ago, # |   0 almost got the idea of DIV2 C :(
•  » » 3 years ago, # ^ | ← Rev. 2 →   +11 It's really interesting who was the author. I will say that it's really interesting to know which problem was made by whom. Golovanov399, could you please show us!
•  » » » 3 years ago, # ^ |   +1 YES! Please Golovanov399! please!
•  » » » » 3 years ago, # ^ |   +57 The author of Div1A and Div1C is Golovanov399, I am the author of all other problems.
•  » » » » » 3 years ago, # ^ |   +11 Thank you!
 » 3 years ago, # |   +62 div2 D was way easier than div2 C !! anyways great contest indeed !!
•  » » 3 years ago, # ^ |   0 Can't agree with this. I solved Div2 C in less than 15 minutes, and couldn't solve Div2 D at all. IMHO, the second part of solution in editorial is more difficult than it could be, because after sorting the pair array, we can simply use two pointer method
 » 3 years ago, # |   +83 I think proof of Div1 D looks incomplete. You should also consider a case when diameter and optimal path intersect, I think they are a bit different.
•  » » 3 years ago, # ^ |   -155 Please mark your comment with spoilers. Some participants might want to upsolve without spoilers.
•  » » » 3 years ago, # ^ |   +411 Well, reading comments under editorial is a bit reckless if you want to solve problems without spoilers
•  » » 3 years ago, # ^ |   +20 Yeah, little bit.
•  » » 3 years ago, # ^ |   0 Correct me if I'm wrong, but I think it still holds when $C = D$ (optimal path intersects diameter) and when $D = F$ (optimal path begins somewhere on diameter), so that should cover all the cases. The proof does not make assumptions on edges existing between nodes in the diagram, so setting paths have zero length should be fine.
•  » » » 3 years ago, # ^ |   0 If C=D then everything is fine, but if diameter and considered path have some edges in common then it is a bit different
•  » » » » 3 years ago, # ^ |   0 ah, i see. i agree that it's incomplete then, there's no way to cover that without another case.
•  » » 3 years ago, # ^ |   0 I'm guessing most people figured out the proof for that case by themselves, but I'll share mine for the sake of completeness (sorry for necroposting lol)Firstly, define $p(U, V)$ to be $1$ if the parity of the number of stone roads on the path from $U$ to $V$ is odd, and $0$ otherwise. Also, define $|UV|$ as the length of the path from $U$ to $V$.Once again, suppose that $AB$ is a diameter of the tree, and $EF$ is a maximum-length path with an even number of stone roads.Let's first show that $p(A, F) \neq p(E, B)$. There's two cases, depending on $p(C, D)$.Case 1: $p(C, D) = 0$Since $p(A, B) = 1$, it must be true that $p(A, C) \neq p(D, B)$. Since $p(E, F) = 0$, it must be true that $p(E, C) = p(D, F)$. The condition holds.Case 2: $p(C, D) = 1$The same thing happens, but the equality and inequality flip places. $p(A, C) = p(D, B)$ and $p(E, C) \neq p(D, F)$. In this case, the condition holds as well.Without loss of generality, suppose that $p(A, F) = 0$. It must be true that $|EB| \leq |AB|$, since the latter is a diameter. Hence $|EC| \leq |AC|$ and thus $|EF| \leq |AF|$. Once again, we've constructed a valid path which is at least as good, and has an endpoint at either $A$ or $B$. This is a contradiction.
 » 3 years ago, # | ← Rev. 2 →   -50 Golovanov399 why my sol for div2 D getting tle https://codeforces.com/contest/1435/submission/96683823 please help...
•  » » 3 years ago, # ^ |   +18 Try taken.lower_bound() instead of lower_bound(all(taken))
•  » » » 3 years ago, # ^ |   0 Why does this work?
•  » » » » 3 years ago, # ^ |   +13 In simple words lower_bound(all(v)) uses random access iterators and works well only for vectors, arrays etc. whereas s.lower_bound is a function built to handle binary search for sets, multiset etc where elements cannot be randomly accessed. there is a blog about this, try searching if interested further.
•  » » » » » 3 years ago, # ^ |   0 Do you have that link?
•  » » » » » » 3 years ago, # ^ |   0
•  » » » » 3 years ago, # ^ |   0 In c++ set or multiset, it is better to use the in-built lower bound function as in the worst case, it has logarithmic complexity. The other one has a quadratic worst case time complexity.
 » 3 years ago, # |   +52 Is tourist's submission for problem E somehow rejudged?
•  » » 3 years ago, # ^ |   +6 Yeah, it was stated that $v_i > 0$, while in fact it was generated (and then validated) that they can be zeroes. It turned out that the second Gennady's submission failed just because of that (and then he wasted the last hour finding out what was wrong).
•  » » » 3 years ago, # ^ |   +8 He came first without having to solve D, so it is all good. Otherwise, the round would have been unrated for him right?
•  » » » » 3 years ago, # ^ |   -8 I guess yes.
•  » » » » 3 years ago, # ^ |   +6 He doesn’t need D and he can be the first.
•  » » » » » 3 years ago, # ^ |   +10 If someone had solved all problems, I'm pretty sure tourist wouldn't have come 1st. If the statement was correct initially, he would have probably spent the last hour on solving D maybe.
•  » » » 3 years ago, # ^ |   +117 What a good way of preventing someone from AKing your contest. (Just kidding. )
 » 3 years ago, # |   +27 There's O(n) solution for div2 D using stack. See my code.But it's hard to explain for me..
•  » » 3 years ago, # ^ |   +5 Wow! That's an amazing solution! Thanks!
•  » » 3 years ago, # ^ |   +5 Great! Thanks for the code
 » 3 years ago, # |   0 Did anyone solve D2C with DP? Is it even possible?
•  » » 3 years ago, # ^ |   +3 I did and got FST :(
•  » » 3 years ago, # ^ |   0 Yes I tried Div2 C with DP, but it failed on test-case 25. But I realise now that it is not possible to solve it using DP. Here is my submission 96694460Atleast in the way I have defined my DP states.
•  » » » 3 years ago, # ^ |   0 Same pretest got us :(
•  » » » 3 years ago, # ^ |   0 I believe my solution has the same idea 96688698, but why do you think it's not possible?
•  » » » » 3 years ago, # ^ |   0 Because in DP solution answer of current state is dependent on previous state and we minimise the difference between minimal and maximal fret. But consider the 25th Test case. 101 146 175 if we subtract 1 96 100 respectively from it we obtain our answer.But DP solution will choose :- 96 96 100 for subtraction, because difference between 5 and 50 is smaller as compared to 100 and 50.
•  » » » » » 3 years ago, # ^ |   +2 Lol, I combined our solutions and it got AC. That's weird. I also added sorts to my solution as you did 96712200
•  » » 3 years ago, # ^ | ← Rev. 2 →   +5 I have a DP + Greedy solution. Here's my submission.Explanation — First comes the greedy part. By observation, it can be seen that after sorting both the notes and $a_{i}$, it will always be optimal to first choose some elements from the first string, then some from the second, and so on. So, after choosing a particular string $k$ for some $i-th$ note, the $(i+1)-th$ note will be always from a string $j\ge k$. This can be implemented using DP.In my solution, $dp_{i,k,j}$ is the state where the $i-th$ note is being considered to be assigned string $j$ and the $(i-1)-th$ note has been assigned the string $k$. The second.first, second.second and first fields of the states represent the minimum value chosen till now, maximum value chosen till now and their difference respectively.
•  » » » 3 years ago, # ^ |   0 Thanks very much for the code. Codes speak for themselves
 » 3 years ago, # |   0 Was able to solve B today but not A :|
•  » » 3 years ago, # ^ |   0 Me too, feels really bad. Another bad thing is stuck on C but then find D so easy after contest
 » 3 years ago, # |   0 Div2C : Each note can be played on a specific range of frets. So out of all such ranges, find the minimum end point (at least one note cannot be played using a higher numbered fret) and the maximum start point (at least one note cannot be played using a lower numbered fret) and output max(0, max_start — min_end). Can anyone point what is wrong with this approach and provide a simple counter test?
•  » » 3 years ago, # ^ |   +4 Firstly, it would not work on test: Test #11 1 1 1 1 5210 13 AnswerThe best solution is to choose 1 note for 10 and 5 for 13. So, the answer will be 1, but your code gives 0.And also consider this testcase Test #21 1 1 1 1 536 8 10 Answermax_start = 5 (10 — 5), min_end = 5 (6 — 1). Your code will give 0, but you cant get 5 from 8 (only 3 and 7).I hope I understand you correctly.
•  » » » 3 years ago, # ^ |   +1 Thanks! Got it.
•  » » » 3 years ago, # ^ |   +1 k1ps thanks a lot for these test cases : )
 » 3 years ago, # |   0 can someone tell my why I am getting tle https://codeforces.com/contest/1435/submission/96685352 in problem D.
•  » » 3 years ago, # ^ |   0 Your solution is O(N^2).final.size() will become N gradually. Since the outer loop is running 2*N times, complexity is O(N^2).Note: 1+2+3+...N = N(N+1)/2 = O(N^2)
 » 3 years ago, # | ← Rev. 3 →   0 Problem D2C/D1A has a DP tag in it. I would really like someone to help me with the DP solution for D2C/D1A if it is possible. Thanks in advance !
•  » » 3 years ago, # ^ |   0 I have shared my DP + Greedy approach here.
 » 3 years ago, # | ← Rev. 2 →   0 i got TLE with N*M complexity on B, helllooo? what's wrong with this " https://codeforces.com/contest/1435/submission/96705114 " ??? i got accepted with " https://codeforces.com/contest/1435/submission/96705022 "
 » 3 years ago, # |   +4 Super easy solutions and implementation for problems Div2 D and Div2 E: Div2 D: 96705496 Div2 E: 96705529
•  » » 3 years ago, # ^ |   +4 D is much easier that the one u linked, just one stack and O(n).
•  » » » 3 years ago, # ^ |   +4 Yes, I see. Complexity can be further reduced to O(N).
•  » » » 3 years ago, # ^ |   +3 Thanks for the solution, mate! Very nice observations
 » 3 years ago, # | ← Rev. 2 →   0 In problem Div2D, if we are getting queries of form "— num", we are updating the lower bound , what we need to do when we get query of type "+" ? I didn't understand this line from editorial "and remove any one of them, because we cannot remove any other shuriken"
 » 3 years ago, # | ← Rev. 3 →   0 Question D div2 can be solved in O (n) rather than O(n*log(n))with simple stack and easier implementation, here is an ac solution. https://codeforces.com/contest/1435/submission/96693538(Only the solve1() function is the code for this qn)
 » 3 years ago, # |   +9 4 + - 1 + + - 4 + - 2 - 3I have hacked my own solution for Div 1 B/Div 2 D with the above test case. My submission:- 96706130 Systems tests are so weak Golovanov399 amethyst0 Endagorion AndreySergunin
•  » » 3 years ago, # ^ |   +26 Wow, I'm really surprised by that considering how easy it is to make good YES tests: just make a bunch of small random YES tests and it's really easy to combine them into 1 big test(just increase the numbers in each small test by the sum of $n$ of the small tests before it)
•  » » 3 years ago, # ^ |   -42 each price from 1 to n occurs exactly once
 » 3 years ago, # | ← Rev. 2 →   +52 Div2 D O(n) solution explanation: note that if we have several consecutive -x, they must be in an ascending order (otherwise we say NO and stop instantly) if we have + and a -x after it, we can collapse them (it's not hard to understand that if the answer is YES, we can act this way, just a simple mindfulness exercise) Bingo! We've just solved this problem in O(n) using stack (going in the reversed order, putting goods on a stack and then taking them from the top of the stack and putting on the showcase). The only tricky moment left: you must pay attention to the amount of shurikens to avoid the lack of them (e.g. example 2 from the problem).
 » 3 years ago, # |   0 Someone please tell where I went wrong 96696335
•  » » 3 years ago, # ^ |   0 fixed
 » 3 years ago, # |   +1 Can anyone tell me why my submission for Perform Easily failed pretest 9 : 96698325 ?My logic :The maximal and minimal frets are obviously useful frets i.e. they are used by some fret. So I first sorted the notes and then fixed the "minimal used fret" by using the frets required by the first note by the various strings (the logic being that the smallest note will use the smallest fret from all possible combinations). So out of the 6 frets usable by the first note, one of them should be the "minimal" fret of the optimum set. Then I just binary search for the right hand side by checking if all notes can be played for the specified minimal fret.
 » 3 years ago, # |   +118 Div1 D can be solved via top trees without analyzing the problem.
•  » » 3 years ago, # ^ |   +44 It seems to be difficult to find any information regarding top trees. Do you have any article regarding this?
•  » » » 3 years ago, # ^ |   +36 At a high level, top trees are just link-cut trees but with a BBST of light-children at each vertex, which allows you to do subtree queries. If you don't understand link-cut trees, learn those first.The main paper for the popular splay-tree-based top trees is "Self-Adjusting Top Trees" by Tarjan and Werneck. This also contains references to alternate works by Frederickson and others which have alternative (but more complex) constructions, so you can read those for more background/theory.I submitted a top-tree solution to this problem 96759325, but it's rather complex and not a great way to learn. The quintessential top-tree problem is SONE1 at http://www.lydsy.com/JudgeOnline/problem.php?id=3153, but the site seems to be down right now.
•  » » » 3 years ago, # ^ |   +8 https://negiizhao.blog.uoj.ac/blog/4912Here is a brief introduction of top trees (with some applications I could come up with), but it is in Chinese. I may supply an English version later.
 » 3 years ago, # | ← Rev. 2 →   0 How to solve Div 2D using priority queue ?? I don't understand this line from editorial. Otherwise, for all shurikens that had a lower bound of something less than x we increase it to x, and remove any one of them, why do we have to remove any one of them when we exactly know which one to remove , how to handle the other type of operation "+" type .
•  » » 3 years ago, # ^ |   0 You can check this solution 96682108
•  » » » 3 years ago, # ^ |   0 can you please explain the idea behind your solution .
•  » » 3 years ago, # ^ | ← Rev. 2 →   +11 We can iterate an array in a reverse order. (Example 1) + + - 2 + - 3 + - 1 - 4 we can create an array like this [-1,-1,2,-1,3,-1,1,4] where -1 indicates that the element was inserted(+ sign). Now maintain a min-heap. Iterate backward on this array, we will keep on inserting the elements. We will first insert 4 and 1. Now whenever we land on -1, it means at this point, some shuriken must have been added on the showcase) which can only be either 4 or 1(Since they are sold at the last). So, we can safely remove the minimum element from the heap and add it to the answer. If the heap is found empty at -1 or if we find one of the element in the heap is less than the element to be insert then the array is inconsistent. Also, we can check if number of +'s is equal to n. If it is not equal then the answer is "NO". 96748102
•  » » » 3 years ago, # ^ |   0 Hey, can you elaborate this part we find one of the element in the heap is less than the element to be insert then the array is inconsistent. ?
•  » » » » 3 years ago, # ^ | ← Rev. 3 →   0 The element x which is currently being added indicates the element was sold at this moment and the elements in the heap stores the element already sold out. It also means that at this point, the elements which we have on the showcase are x and all the elements in the heap. By what question says, x has to be the minimum as the customers purchases the minimum element on showcase, but in this case, there is some element in our heap which is less than x and the customer have purchased x instead of that minimum element in the heap. So, it is inconsistent.
•  » » » 3 years ago, # ^ |   0 Thanks for the explanation. Can you please explain how is this safe? we can safely remove the minimum element from the heap and add it to the answer
 » 3 years ago, # |   0 Why my submission link1 getting TLE but when i changed vector size to (n*m + 5) it get accepted link2 ?
•  » » 3 years ago, # ^ |   +5 This is a common mistake when there are multiple testcases. It may take 500*500*t if you create/initialize 500*500 arrays every time, and that's about $2.5 \times 10^{10}$. The statements said that "Sum of $nm$ over all test cases does not exceed $250000$" so the second submission could pass.
 » 3 years ago, # | ← Rev. 2 →   +10 About Problem 2E/1C: I didn't like this problem at all. It was boring and a detail-finding kind of problem where you know right after reading the problem that you can get to the solution if you are not lazy enough.about the $-1$ part, we can just see that since $a>bc$, for each spell, whether complete or not, contributes positively to the damage. Hence the damage can be arbitrarily large. No need to think about overlapping parts, as is done in the editorial.
 » 3 years ago, # |   +13 Linear time solution of div1B: Consider the queries backwards. Then — x actually means adding shuriken with value x to the showcase, and + means deleting some shuriken. So we can maintain the sorted stack of all shurikens at the moment. On + we will delete the minimum from the stack (it's easy to see that it's always the best option), i.e. top element, and add it to the answer array. On — x we will check that x is less then the value on top of the stack and add it. If we successfully looked through all the queries we will just print the reversed answer array, and otherwise there's no answer. My code 96667703
•  » » 3 years ago, # ^ | ← Rev. 3 →   0 I had a similar idea , which unfortunately didn't work. can you please explain what's the flaw with with the following logic.Initialise cnt to zero . While processing the queries backwards , if we see + we increment the cnt variable by 1. When we see — x , we push x in the stack if x is less than top element otherwise we can pop atmost cnt number of elements from top of the stack such that now either the stack is empty or top of stack is greater than x.while popping from the stack we decrease the cnt by 1.
 » 3 years ago, # |   0 There is something I can not understand in the proof of 1D.Tutorial says "Let AB be a diameter of the tree, and the optimal answer be EF." But it seems that in the given tree, AB is not a diameter.Did I understand what the tutorial wanted to express wrongly? Could anyone explain it for me? Thank you in advance!
•  » » 3 years ago, # ^ | ← Rev. 2 →   0 This is for the sake of brevity. You can see that the edge CB is a little bit longer than AC. Apparently that means that there are some nodes on a path between B and C
•  » » » 3 years ago, # ^ |   0 Ha, got it! Thank you sir!
 » 3 years ago, # |   0 can someone explain to me whats wrong with my algorithm 96738918 in D2C? i know my idea is wrong,but im confused whats wrong with it
•  » » 3 years ago, # ^ | ← Rev. 2 →   0 The answer may not always be abs(b[n]-a[6]-b[1]+a[1]). An example: 1 3 6 8 100 1000 3 1003 1250 1500 According to your code, the answer must be 502, but the correct one is 398.
 » 3 years ago, # |   0 Can someone explain A?
•  » » 3 years ago, # ^ | ← Rev. 2 →   +3 Hint1N is even .. So try to make every two element operation sum = 0 Hint 2Suppose think there have two number a and b.1. a=2 & b=3 both positiveThen answer is b=3 & -a=-22. a=2 & b=-3 negThen answer is -b=3 & a=2.3. a=-2 & b=-3then answer is b=-3 & -a=24. a=-2 & b=3then answer is -b=-3 & a=-2 If have any query..please ask.
 » 3 years ago, # | ← Rev. 2 →   0 Can someone please tell me why my solution is wrong for Div2. A... 96650687I printed all 1's if the sum of all elements of original array is 0 and I got wrong answer on 10th Test.
•  » » 3 years ago, # ^ |   0 if the sum ever does happen to be 0, then you're printing 2 valid answers which is why you're getting WA. Just add a return statement.
•  » » » 3 years ago, # ^ |   0 Got it! Thanks
 » 3 years ago, # |   0 Could anyone recommend similar problems to div2C&|D?
•  » » 3 years ago, # ^ |   0 Just search the tags
 » 3 years ago, # |   +1 How to approach the problem E with ternary search( Proof of maximal value being a convex function) ??
 » 3 years ago, # |   +1 Was div.2-D a standard problem? Everybody is saying that the problem is very easy but I could not come up with a simple idea for the problem. I was thinking of a heavy implementation solution.
•  » » 3 years ago, # ^ | ← Rev. 2 →   +1 The idea of problem D:- 1) pair every "+" with below "-" using upper bound. 2) check the order of elements with given order using stackfor better understanding, check out my code
 » 3 years ago, # |   0 I think there's also a O(n) solution for Div2 D using list and it's easier to explain.We consider matching the '-' operations with '+' operations from "- 1" to "- n".After we matched a '+' operation and a '-' operation, we remove them from the operation sequence.So, when considering the operation "- x":If the previous operation is "-", it's impossible to meet the requirements because we have already removed all the numerically smaller "-" operations.Otherwise the previous operation is "+" and we can easily match this "-" operation with it.This can be maintained with a list by recording the position of each "-" operation in the operation sequence.My code
 » 3 years ago, # | ← Rev. 2 →   0 All problems(DIV2) are good and do not need a hard data structure. some observation and you can solve.
•  » » 3 years ago, # ^ |   0 Except the one about ramen
•  » » » 3 years ago, # ^ |   0 I was talking about DIV2
•  » » » » 3 years ago, # ^ |   0 Oh i was doing the original Technocup and there we had that problem about ramen, so i thought it was in div2 too
 » 3 years ago, # | ← Rev. 3 →   +1 Div2 D can be solved going through the reversed query with a set. Every time we meet a "+" we can just say that the woman takes the one with the least price back. So if set's size is 0 the answer is NO. Else erase that element and add to the answer. Else if we meet a "- x" then x must be the less then the least element in the set now. If not, the answer is NO. If yes, insert x. 40 lines of code.p.s. reverse the answer in the end96676641
 » 3 years ago, # |   0 Can anyone tell me why am i getting TLE in div2B? An n^2 solution should have been fine ,right? 96669009
•  » » 3 years ago, # ^ | ← Rev. 2 →   0 I think n^2 should be too slow (2,5e5)^2 is a lot
•  » » » 3 years ago, # ^ |   0 Oh. Can you kindly explain your approach for B then?
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   0 Oh i see the problem. Im sorry thought it was (nm)^2, but after seeing ur code i saw it's nm, which is ok. The problem must be creating a huge array vis, since all other things have a good time complexity.p.s. I got ur sol accepted with changing vis size to n * m + 1
 » 3 years ago, # | ← Rev. 2 →   0 I don't know why I TLE in B :(( somebody help me, please!!!!!!! MY SUBMISSION
•  » » 3 years ago, # ^ |   0 You used memset for 1000000 bytes in while loop for 100000.
 » 3 years ago, # |   0 https://codeforces.com/contest/1435/submission/96662042 can someone tell me what I am wrong?? I can't find anything wrong with my code. please...
•  » » 3 years ago, # ^ |   0 the first matrix given gives you the column of the number and the second matrix gives you the row of that number
 » 3 years ago, # |   0 How to solve div2E with binary search?
 » 3 years ago, # |   0 How could anyone in A saw the row of array like: -a2, a1, -a4, a3, ..., -an, an-1. I mean, no advices on that path weren't given, is anyone sorted out how this works?
•  » » 3 years ago, # ^ |   0 n is even, and you want to make two adjacent elements make a sum of 0.if you have the array a1, a2 — make the sum a1*(-a2) + a2*(a1) = 0.If you do it to every adjacent pair, the total sum will be 0.
•  » » » 3 years ago, # ^ |   0 Yeah, I understand that I need sum = 0, but I hadn't sorted out how to achieve this at the competition, so that's quite bad. That's pretty hard for me to find easiest ways to solve a problem.
•  » » » » 3 years ago, # ^ |   0 Emmm. What you need to do is just practicing more and thinking more. I believe you can solve such problem quickly in a nearly future.
 » 3 years ago, # |   +5 You can solve KOIREP from SPOJ if you have solved Div 2C/1A
 » 3 years ago, # |   0 Could someone please explain the solution of the div1E？I'm totally confused with the conclusion “the outcome of the game is defined by the index of the last move and the last difference between the elements”and“ if we fix the last index and gradually decrease the last difference, the grundy value will not decrease. ”
 » 3 years ago, # |   0 Why we used set>> in div2C why can't we just keep track of the value(0-5) in an array of n
 » 3 years ago, # | ← Rev. 2 →   0 Could anyone please tell me some other solutions (except two pointers) of the problem Perform Easily? There are so many problem tags for this problem, so I guess there will be some interesting solutions (such as binary search).
•  » » 3 years ago, # ^ | ← Rev. 2 →   0 I thought of a different approach and used lower bounds and upper bounds but it is failing at test case 9 can you tell me whats wrong with my code? you can go through this I posted this earlier Can Someone tell me why I my code fails at test case 9 96809688 I have used the approach that after sorting the notes the there cannot be a fret smaller than difference between first note and first string then I took two case 1. I found all the frets for every note which is just smaller or equal to the smallest fret. 2. I found all the frets for every note which is just larger or equal to the smallest fret. then I found the Min of differences obtained from these two.
•  » » » 3 years ago, # ^ |   +1 Sorry, I don't understand your approach. However, there is a set of small data that your code can't get the right answer, maybe you can try to discover your mistakes through it.Input1 1 1 96 99 1003101 146 175Output50Your answer74Good luck!
 » 3 years ago, # |   0 5 minutes left time to be ready : )
 » 3 years ago, # |   0 Thx for this round!