### awoo's blog

By awoo, history, 7 weeks ago, translation, 1680A - Minimums and Maximums

Idea: BledDest

Tutorial
Solution (BledDest)

1680B - Robots

Idea: BledDest

Tutorial
Solution (BledDest)

1680C - Binary String

Idea: BledDest

Tutorial
Solution (BledDest)

1680D - Dog Walking

Idea: vovuh and BledDest

Tutorial
Solution (vovuh)

1680E - Moving Chips

Idea: vovuh

Tutorial
Solution (vovuh)

1680F - Lenient Vertex Cover

Idea: BledDest

Tutorial
Solution (awoo)  Comments (42)
 » Can someone share his sliding window approach for problem C ?
•  » » Reduce the problem: find the lowest cost substring, cost = max(0's, total 1's — 1's)The cost of the final optimal substring will either be contributed by 0's or by 1's. Let's binary search on the best answer that can be contributed by 0's and then binary search again on 1's, then we take the minimum of the two. To check if an answer g is possible, we run sliding window. For example, let's say we are currently binary searching on best answer contributed by 0's. We can expand the window until we have more than g 0's in the current window. Then we can shrink it from the left until we have exactly g 0's again. here's my submission: https://codeforces.cc/contest/1680/submission/157099290
•  » » » This will work when we even check the contribution of just ones or just zeros, right? Please confirm.
•  » » » No need to check for 0's, we can show that if some cost k is achieved we can achieve it using the number of 1's removed = k (and hence the number of 0's left <= k). The proof of this is, suppose on the contrary we get a score of k and remove less than k 1's, then the number of 0's left in the string is k. Now we should not be able to remove more 0's from the string, or we will get a lower score. Hence there should be sufficient 1's surrounding the 0's so that we can bring the count of removed 1's up to k.
•  » » Sliding window / two pointers approach without binary search, runs in O(n) time, in C. Let our window represent the string left after deleting the first i elements, and the size - j elements from the right. We know that for a window starting at index i, if we're increasing the window size, then the best cost we can get is when the number of characters 0 in our window is equal to characters 1 outside the window. This is because if we keep increasing our window size then the number of character 0 in the string will keep increasing, and a smaller window will increase the number of character 1 outside the string. Then we just iterate through i, the beginning of each window, and keep the maximum window we can get. When i goes to the next iteration, it will find the next possible window starting at the previous i's max (to keep it O(n)), while also keeping track of the minimum cost as well.
•  » » » Great approach !! Understood completely. Thx
 » Can someone share all the approaches to question C as mentioned in the tutorial? It will be nice to see them all at once and compare their complexities Dynamic programming, Greedy, Two pointers
 » Has anyone solved problem C with dynamic programming？
•  » » Did you ?
•  » » dynmaic programming(prefix sum) - Submission
•  » » » Can you please elaborate your solution. Precisely what the last for loop means.
•  » » » » Refer to this COMMENT for explanation.
 » 7 weeks ago, # | ← Rev. 2 →   Can C be solved using ternary search? If yes, can someone share their solution for the same?
•  » » 7 weeks ago, # ^ | ← Rev. 3 →   Here is my solution with ternary search.157125272I think the solution can be improved and may be done more clearly.
•  » » » I also tried it using ternary search but got WA as verdict , can you help me finding the test case where it is failing ? 157066350 For every index j in the string , I am finding the optimal i so that cost is minimum using ternary search .
 » Binary Search Video Solution for C
 » green !
 » I used a segment tree.I think I can replace it by a monotune queue.
 » Problem F looks very similar to https://codeforces.cc/contest/19/problem/E
 » For problem E: a "state machine" approach works, without any DP.The idea: same as in the editorial, we sweep all the chips from left to right. If we reach a column with a single chip, just move it to the right and add 1 to the answer. What if we reach a column with 2 chips (whether it's 2 chips that are already there, or 1 chip is there and another has been pushed in from the left)? In this case, we push the top chip down or the bottom chip up. But we don't have to consider both possibilities yet. Just put the chip in an "indeterminate" state and resolve the state when you reach a column with exactly one chip.Of course, this doesn't save much in terms of runtime/space — the dp solution only carries around 2 integers. But I think it's a nice way of looking at the problem.
 » Hey, can anyone help me with problem C. I've thought of a greedy solution whose basic idea was to remove the max number of zeros by removing min number of ones and repeat this till the string becomes empty from zeros.I tried checking with many edge cases but cannot see what is the mistake ive made. So plz can anyone help me I have commented the code for better understanding https://ideone.com/bBoj0b
•  » » Failing testcase: Ticket 7173
•  » » » Thanks!! I didnt know there was some website like this
 » I think there is a typo in the editorial of problem E, because the transition $dp_{i, 1} \rightarrow dp_{i+1, 1}$ is considered twice. It's not really important though because it's easy to understand that the last transition is meant to be $dp_{i, 1} \rightarrow dp_{i+1, 0}$
 » If you are/were getting a WA/RE verdict on problems from this contest, you can get the smallest possible counter example for your submission on cfstress.com. To do that, click on the relevant problem's link below, add your submission ID, and edit the table (or edit compressed parameters) to increase/decrease the constraints. A: Minimums and Maximums B: Robots C: Binary String D: Dog Walking E: Moving Chips F: Lenient Vertex Cover: Will add if anyone's stuck on debugging (only till the next 7 days). If you are not able to find a counter example even after changing the parameters, reply to this thread (only till the next 7 days), with links to your submission and ticket(s).
 » Finally an expert after this contest, really liked third problem
 » 7 weeks ago, # | ← Rev. 2 →   i really want to solve the problem c by myself but i am stuck for a few days. can anyone give me any insight to approach the problem
 » In problem C, I iterate to [0,n-1] taking only the 'i'th character at each iteration as the first character of the substring, and for finding the last position of the substring, use the binary search. so how do we binary search it! first, compute the prefix of both '0's and '1's in the string, then use those prefixes to compute the max('0's present in the substring, '1's deleted from the substring). if('0's present in the substring > '1's deleted from the substring) then shift left=mid+1; else, shift right to mid. And then compute the minimum of answer and the substring computed up to left. https://codeforces.cc/contest/1680/submission/157425332
 » 7 weeks ago, # | ← Rev. 4 →   Problem F is very nice.The final part can also be done with LCA. If you want to remove an edge in the tree, which splits it into two subtrees, the edge needs to be so the bad edges not in the original tree (ie. those non-tree edges that connect vertices of the same color) need to be in separate subtrees so we can flip the colors of one subtree.Then the problem becomes: Given several paths on a tree, find an edge that lies on all of them. This will be the edge we remove. The intersection of two paths is empty, a vertex, or a subpath. It can be checked with a few cases using LCA whether the intersection is another path, and we repeat checking with this newer subpath.
•  » » Hey, do you think you could help me understand solution to problem F? I understand when it says we need to remove one edge such that the remaining graph is bipartite but I don't after that I don't really understand anything about how we find this edge. Thanks.
 » 7 weeks ago, # | ← Rev. 2 →   I don't understand how the dog walks in the second example of question D
•  » » Now I understand，can't ignore the order of numbers
 » 7 weeks ago, # | ← Rev. 2 →   Can anyone help finding which TC my submission fails on? https://codeforces.cc/contest/1680/submission/157535516Edit: Problem E
•  » » I did the same! IDK where I am going wrong. If you get it do tell me.
•  » » Failing testcase: Ticket 7267CC: Agent_Lemon
 » Can someone please hack my program for D? Thank you in advance. https://codeforces.cc/contest/1680/submission/157643976
 » 6 weeks ago, # | ← Rev. 3 →   I found a different simple solution for D.Let $p[i] = \sum_{k \leq i} a[i]$ be where the dog is after $i$ steps.Let $pre[i]$ denote sum of nonzero (known) steps in the first $i$ steps.Let $cnt[i]$ denote the number of zeros in the first $i$ steps.The problem is equivalent to maximizing $\max_i p[i] - \min_i p[i].$ Thus, for each pair of indices $(i,j)$ we check if we can make $p[i]$ the min and $p[j]$ the max or vice versa, and what optimal difference we can get with this pair. (Clearly if $a[i] > 0$ then it cannot be the min so we can ignore such cases, etc.)We need the following inequalities to hold: $p[i] \in [pre[i] - cnt[i] \cdot k, pre[i] + cnt[i] \cdot k]$ $p[j] \in [pre[j] - cnt[j] \cdot k, pre[j] + cnt[j] \cdot k]$ $p[j] \leq (cnt[n] - cnt[j]) \cdot k - (pre[n] - pre[j])$ $p[j] - p[i] \in [(pre[j] - pre[i]) - (cnt[j] - cnt[i]) \cdot k, (pre[j] - pre[i]) + (cnt[j] - cnt[i]) \cdot k]$ where the first two inequalities are obvious, the third inequality ensures that the dog can return to zero after reaching the maximum at $j,$ and the fourth inequality is that the dog can go from the min at $i$ to the max at $j.$It can easily be checked whether these four inequalities hold, and find the optimal difference for this pair $(i,j).$ The overall runtime is $O(n^2).$
 » Here's a funny solution to E using graphs.Form a graph where vertices are the chips. Initially we can connect all pairs of edges between the vertices, where distance between two vertices is the manhattan distance. Then the answer is just given by the MST of the graph: Moving chip $i \to j$ corresponds to taking that edge. Finally, notice that for any chip, we only need to consider edges from that chip to its (at most) four nearest neighbors. The ones closest to the left and right of it in the same row, and similarly for the other row. This is because if we move the chip to any of the other chips in a move, we can improve this by first moving to one of these four chips along the way and the number of moves is the same. Thus the number of edges in the graph is $O(n),$ so MST can be done in $O(n \log n).$
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   Unfortunately, the MST is not the answer because we can combine chips at intermediary positions. Intuitively, a counterexample in euclidean space would be a triangle (the sides of the triangle would be the MST but we can move the chips to the middle).A counterexample in the context of the problem would be:.*.*...*.*The MST would take 6 moves, but they can be combined in 5.
 » Hello, BARBARIANNNNN , can you please explain your implementation of 157031361, i.e. problem 1680C - Binary String , i am not getting a proper intuition of your code. And can you please tell me this condition : while(j<=i && c0>x) { } // my doubt is why we are only comparing c0 with x and not change the loop like this while(j <=i && c1 > x) { } if(c0 <= x && c1 <= x) return true; changing this is actually giving the wrong answer and giving the bigger cost . Why is it so as we are also considering all the cases in comparing c1 with x ?
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   In the function check, we use $[i,j]$ mention the subsegment after our operations. Because we want the max of the number of characters 0 left in the string and the number of characters 1 removed from the string to be $\le x$, we should control the c0(the first number in the condition) $\le x$, while we check whether c1(means the second number in the condition) also $\le x$. If we change the loop like while(j <=i && c1 > x), then when you add j, the c1 will increase, and it is meanless(never c1<=x because of increasing).
•  » » » Thank you very much BARBARIANNNNN for taking out your time and expalining it clearly. Actually, TBH as a newbie i do not expect so much replies from high rated coders as i am a newbie. But when someone takes out time to resolve my doubt, it just increases my enthusiasm , makes me happy and fills me with a new vigor to practice more. Really, Thank you again for your time to resolve my doubt BARBARIANNNNN as i have been actually stuck for a long time trying to understand the intuition of your implementation.