Vovuh's blog

By Vovuh, history, 2 months ago, In English,

1029A - Many Equal Substrings

Tutorial
Solution (Vovuh, O(n^2))
Solution (Vovuh, prefix-function)

1029B - Creating the Contest

Tutorial
Solution (Vovuh)

1029C - Maximal Intersection

Tutorial
Solution (PikMike)

1029D - Concatenated Multiples

Tutorial
Solution (PikMike)

1029E - Tree with Small Distances

Tutorial
Solution (Vovuh)

1029F - Multicolored Markers

Tutorial
Solution (PikMike)
 
 
 
 
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2 months ago, # |
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May I know how to solve C using segment tree? A lot of folks seem to have done it this way.

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    2 months ago, # ^ |
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    You can use segment tree to get range min and range max instead of doing a clever precomp. This adds log factor. I did it with RMQ table because I am lazy.

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    2 months ago, # ^ |
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    You can use multiset to get O(n log n) with a simpler solution: 42096347

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      2 months ago, # ^ |
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      yes it is simple and elegant and effective.

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      2 months ago, # ^ |
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      Brilliant!

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      2 months ago, # ^ |
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      Your solution looks really small and pretty. Can you explain the logic behind it ? I can't seem to get it.

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      2 months ago, # ^ |
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      Okay i got what you were trying to do. I guess multiset is actually a kind of Self-Balancing BST. Isn't it ? that's why we get n * log(n) solution.

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        2 months ago, # ^ |
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        Multiset works like a set, but allows repeated elements. We just need to find min R and max L, but each time ignoring one of the n intervals. We add all L's and R's to the multisets, and then, for each interval, remove it's values and check for max/min and get best value. It's easy to query because multiset is ordered like set, so min = .begin(), and max = .rbegin()

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      2 months ago, # ^ |
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      This can be used too in 2D like in this recent problem: 1028C - Rectangles

      Solution with multisets: 42187139

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    2 months ago, # ^ |
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    My segment tree answers the question: What is the intersection segment of segments between index l and index r inclusive?

    So, I just iterated over all i that 1 <  = i <  = n, and in each iterate I asked about the intersection segment of segments from 1 to i - 1 inclusive and the intersection segment of segments from i + 1 to n (i.e. I removed the segment with index i), then merged the two resulting intersection segment to obtain the segment which its length should enter in the result maximizing operation.

    For merging two segments [a, b] and [c, d], the resulting segment is [lft, rht], where lft = max(a, c) and rht = min(b, d). Then if lft > rht let this segment equal to [ - 1,  - 1] as a sign for the future merges that this segment contains nothing.

    This is my submission.

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      2 months ago, # ^ |
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      I was actually looking at your solution before you replied! thanks!

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    2 months ago, # ^ |
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    I used Sparse table for RMQ. Whenever I write sparse table, I feel very proud.

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2 months ago, # |
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For problem B,

Can anyone tell me why is my solution failing?

http://codeforces.com/contest/1029/submission/42053121

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    2 months ago, # ^ |
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    Maybe you misunderstood the problem? for every index i, since the list is sorted, you just have to make sure i+1 is less than 2*a[i]. You only need one such value to satisfy the condition. Just replace your search function with a check 2*a[i]>=a[i+1] or something like that.

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      2 months ago, # ^ |
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      Oh yes, Thanks.

      I misunderstood the problem. My bad.

      Thanks again. :)

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2 months ago, # |
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The solution given for problem B, For testcase —
1
1 2 3 4
The solution gives 4 as output, whereas actual answer is 3 [2,3,4].

Edit — My bad, I dint understand the problem :(

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    2 months ago, # ^ |
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    I think we made the same mistake. Can you please check what's wrong with my solution? http://codeforces.com/blog/entry/61439?#comment-454828

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      2 months ago, # ^ |
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      yes,we made the same mistake.
      you are finding only the element which is (<= 2 * a[i])
      we are suppose to check if that element exist and go for next element adding that to the answer
      for suppose —
      1
      1 2 3 4
      what we are doing is —
      at 1 — ans = 2 [1,2]
      at 2 — ans = 3 [2,3,4]
      at 3 — ans = 2 [3,4]
      at 4 — ans = 1 [4]
      our ans is max = 3

      but actually we are supposed to do this way -
      at 1 — check if 2 exists, if so ans = 2
      at 2 — check if any element <= 4 exists if so ans = ans + (index of element — current index + 1)
      and move current index to the element (<=4)
      and keep going so on

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        2 months ago, # ^ |
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        Damn word play!

        "There is only one condition that should be satisfied: for each problem but the hardest one (the problem with the maximum difficulty) there should be a problem with the difficulty greater than the difficulty of this problem but not greater than twice the difficulty of this problem."

        Thanks for the help!

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2 months ago, # |
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Here is one of my solutions for problem E which apparently works in linear time without dynamic programming.

It uses BFS O(n+n-1) = O(n) to get vertexes sorted by distance from the first vertex.

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2 months ago, # |
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I solved the problem A by using Z-function. Also a solution :)

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2 months ago, # |
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The problem E doesn't need any sorting, just dfs on a tree Code

The idea is to connect the parent of the furthest vertexes to the first vertex, and then mark all parent's children as used ones.

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2 months ago, # |
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In D, can't you use a hashmap with the remainders mod k, instead of appending to rem_len_i and binary search? It should decrease the complexity to O(n) too.

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    2 months ago, # ^ |
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    do you mean D?

    yes, you can. here is my (cleaned) solution (which is actually slower than PikMike's) ^_^

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      2 months ago, # ^ |
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      I meant D. Edited thanks! Thanks also for the solution leoniduvk!

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    2 months ago, # ^ |
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    I submitted D 15 times using an O(10 * n * 2) solution (i < j then i > j), finally passed systests with O(10 * n) by precomputing everything first then handling i != j

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      2 months ago, # ^ |
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      Your solution just has the same problem as the majority of all the others, you're doing

      res += map[ x ]

      without actually checking if x was in the map originally. This generates a lot of empty elements that add up to the complexity. Your submission got acccepted with 1980 ms, submit the same code again and it might not even actually get accepted.

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        2 months ago, # ^ |
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        Still TLEs even with the find check

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          2 months ago, # ^ |
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          I know you have checked; what i am saying is that your program still uses a lot of memory, like when i wasn't doing the check, so there's gotta be something wrong.

          So you could run some tests displaying the size of the map(s) before and after inserting some known amount of elements, and checking whether that's correct or not.

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2 months ago, # |
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can someone help me i am getting tle for 1029D — Concatenated Multiples https://onlinegdb.com/rks1jEywm

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    2 months ago, # ^ |
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    if(hash[j].find(temp)!=hash[j].end())
    

    Check this condition ,or else element temp will be created and inserted in map. And changed long long to unsigned long long

    link

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2 months ago, # |
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My solution for F is getting WA for test case #37 42102854. Can someone please take a quick look? Edit: nevermind, found a silly mistake.

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2 months ago, # |
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On problem E i got WA on testcase 11. The answear differs by 1. Can anyone tell why? Code : http://codeforces.com/contest/1029/submission/42104188 Edit: Fixed.

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2 months ago, # |
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The following is a simple and compact C++17 solution for problem 1029A - Many Equal Substrings using the STL function string::substr().

42106199

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2 months ago, # |
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Can someone help me out ! In problem F , #testcase 37: intput: 99999999999973 99999999999971 answer: 199999999999948 we need to find out the minimal perimeter of rectangle formed from a,b. So i calculated the answer for this case is a rectangle has: 3089852923 width and 64728 height. The perimeter of this rectangle is 6179835302 , which is smaller than the answer. So the answer should be 6179835302, not 199999999999948. Edited: my fail , i misunderstand the problem!

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2 months ago, # |
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I solved E using dynamic programming in linear time but it is failing test case 8 , Can someone tell me what is wrong with my code ?

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    2 months ago, # ^ |
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    The answer to this test should be 2. 1 -> 5, 1 -> 8.

    9
    1 2
    2 3
    3 4
    4 5
    5 6
    5 7
    7 8
    8 9
    
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2 months ago, # |
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I can't understand why my solution of problem C fails. Can anyone help me?

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    2 months ago, # ^ |
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    Check this test case:

    3
    10 20
    13 14
    10 10
    
    Answer: 1 
    Your output: 0
    
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      2 months ago, # ^ |
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      Thanks! I've just found the error... It was a such stupid one

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2 months ago, # |
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Can someone explain DP on tree solution for Problem E?

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    2 months ago, # ^ |
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    If you are trying to solve E with DP and are having some WA submissions then I suggest try these following 2 test cases. It is just my humble advice but I did use those 2 cases to find out the AC method.

    tests

    First of all, you can see that we have already had a tree with the root is vertex 1. Since the shortest path from vertex 1 to any other vertex at most 2, you may come up with this idea: let f[u][i] be the minimal number of edges that should be added so that the shortest path from 1 to u is i and the subtree with root vertex is u satisfy the given condition. Thus, every vertex has 2 states: f[u][1] and f[u][2]. However, later you will see that every vertex has 3 states :v

    If the distance from vertex 1 to u is 1, it can easily be seen that (v is a child vertex of u).

    If the distance from vertex 1 to vertex u is 2, vertex u is sure to be connected with a vertex that has the shortest path from vertex 1 is 1. Here now there are 2 options: connect u with its direct ancestor or a child vertex of it. In order to implement, I denote the second subcase as f[u][3].

    f[u][3] = min(f[u][2] - min(f[v][1], f[v][3]) + f[v][1])

    In addition, it can be seen that the base cases are those leaves of the tree and f[u][1] = 1, f[u][2] = 0, f[u][3] = ∞.

    Look through my AC code ?

    Sorry if you consider my English is bad :(

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2 months ago, # |
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Can someone give better explanation for problem D?

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2 months ago, # |
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For D, we can maintain an array tenp for (10i)modk, for 1 ≤ i ≤ 10 (since maximum number of digits is 10). We can also maintain a HashMap, that maps the modulus to it's count. That is the key is (A[i])modK,and the value is the number of times a number in A, that has the same value after moding by k.

Now, for every number A[i], we have to find numbers b such that (b × 10d + A[i])%k = 0.(Where d is the number of digits in A[i]). This can be written as,

((b%k × (10d)%k)%k + A[i]%k)%k = 0

. Let $1$ n = A[i]%k , α = (10d)%k and m = b%k. So, it becomes:

((m × α)%k + n)%k = 0

Thus, if $n=0$, then, we can satisfy the condition when m = 0 or when α = 0. We can search for the number of times a zero is inside the HashMap, for checking the m = 0 condition. If α = 0, then we know that every number int A can be used to satisfy the condition, as the whole number will always be divisible by k.

Now, if n > 0, then (m × α)%k = (k - n), thus,

Thus, we'll just have to count the number of occurrences of $m$ inside the HashMap.

My implementation didn't work quite well, so I think there must be some quirk left behind that may make this implementation unreasonable. Any help appreciated!

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2 months ago, # |
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2 months ago, # |
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For the explanation of question F, what do they mean by "a+b should be area of the outer rectangle." Like is there even an outer rectangle?

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2 months ago, # |
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Problem 1029E Why does this code use set (-d[i], i) instead of (d[i], i)? I used (d[i], i) and failed the test.

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    2 months ago, # ^ |
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    in set it is sorted in increasing order. now in the case of this problem u need to know the furthest distance from the root node. and you need to keep it at the beginning of the set. if you take the positive value of the distance it will be stored at the end. but if you take the negative value it will be the smallest and stored at the beginning. :)

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2 months ago, # |
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In problem D, the only difference between the editorial and my approach is that i've used "unordered_map<int,int> mp[11]" where mp[d][x] stores the frequency of numbers which have d digits and leave a remainder x when divided by k. I get TLE. why?
This is the link to my solution. 42241992

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    7 weeks ago, # ^ |
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    I used an ordinary map, and I too faced the same issue.

    Apparently, checking if an element is present in the map before trying to retrieve it, significantly reduces the running time.

    In your code, instead of

    ans+= mpp[j][y];
    

    do a

    if(mpp[j].find(y) != mpp[j].end())
    ans+= mpp[j][y];
    
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2 months ago, # |
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I think the algorythm for A is incorrect, for 'abab' it gives out the wrong answer — this answer is longer than the right one, can anyone tell me am I right or not? If not, why?

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2 months ago, # |
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For problem A, isn't the first idea in O(n^3)? Because for this example 'aa..aac', for every k we need to verify the prefix in (n * (n + 1)) / 2 steps. But in the first solution, which implements this idea, the author says that it's in O(n^2) 'Solution (Vovuh, O(n^2))'.

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My solution for problem B gets WA on test 4. No idea why. Can anyone help me please? http://codeforces.com/contest/1029/submission/42271885

Thanks in advance!

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7 weeks ago, # |
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In problem 1029C — Maximal Intersection .

Intersection of some segments [l1,r1],[l2,r2],[l3,r3],....,[ln,rn] is [max li,min li].If 
    this segment has its left bound greater than its right bound then the intersection is 
   empty.

Can anyone give me proof of this. I want to know why this happens?

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    7 weeks ago, # ^ |
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    We want to find the longest segment that is an intersection of some segments [l1,r1], [l2,r2], ..., [ln,rn]. To maximize the length, we want to find the leftmost and rightmost points on this segment.

    The leftmost point guaranteed to be in all the segments is max(li) for some i. To prove this, we can use contradiction. Assume the leftmost point guaranteed to be in all segments is lj, for some lj < li and j != i. But then the i-th segment (with endpoints [li, ri]) is not guaranteed to be in the wanted intersection, because a case like rj < li is possible (you can visualize the segments as: [lj]----[rj]....[li]--------[ri]). Therefore the assumption is false and the original claim is true.

    Similar logic for the rightmost point.

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5 weeks ago, # |
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For problem D,can someone tell me why in PikMike's code,he can use code

if (len[i] == j && (rem + a[i] % k) % k == 0) --ans;

to subtract the pair of (i,i)? I think it is possible that there are two numbers which have same length and satisfy the problem but their value don't equal.

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5 weeks ago, # |
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In Problem C, instead of storing result in partial sum like arrays, we can store 4 values which are max of left, min of right and second largest max of l, second largest min of r.

Now, my result is min — max > 0 ? min — max : 0.

So, now I iterate over all the segments and see if removing that improves my answer which maximum length intersection between the remaining.

if the current segments left is equal to max then I replace max my second largest max, else it remains same. Similarly, if current segments right is equal to min then I replace min by second largest min, else it remains same.

Now I keep checking in the loop if my result increases and finally output the maximum result.

This solution also works in O(n) but extra space complexity is reduced to O(1), since it does not require partial sum array.

43075255

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    3 weeks ago, # ^ |
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    Can you please explain why this logic works?

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      3 weeks ago, # ^ |
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      Hi,
      For any given set of line segments,
      intersection length = max(left indices) - min(right indices).

      And for an intersection to occur, this number must be greater than 0.

      Now, if I have to remove 1 segment so as to increase the intersection length, I have to keep track of the new max and min after removing that segment.

      So, basically I will loop over all segments and calculate new intersection length after removing that segment.

      For, instance if I remove a segment whose left index was max, then I have to resort to second max in order to calculate the new intersection length, because that is the new left max now.

      Similarly for min right and second min right.

      Therefore, I have to keep track of max of left, second max of left and min of right, second min of right.

      Since I am removing only 1 segment therefore storing these 4 value will work, whereas if I had to remove k segments then I would have to store more values.

      Therefore, this logic will only work for this particular problem statement only.

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4 weeks ago, # |
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Problem E , my solution is make the edge with vertex v have maximum number of pv (pv is a vertex have distance to vertex v is 1) , why i am wrong . my code : http://codeforces.com/contest/1029/submission/43141795