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# segment tree

 0 Can anyonre give me a hint how to solve this problem here! asked 30 Aug '17, 01:04 94●4 accept rate: 0% An approach for this problem is explained here: http://codeforces.com/blog/entry/10183?#comment-156355. (05 Feb '18, 20:23)

 1 Assuming that you are familiar with segment tree You can solve this problem by building a segment tree, with a sorted sub-array in every vertex and then divide [l, r] into sub-segments, and do binary search on sub-segments. answered 30 Aug '17, 01:19 2★a_d_i 174●6 accept rate: 23% Using sorted sub-array in every vertex will take $n^2 log(n)$ time, no?? If I am getting correctly what you are saying.. (30 Aug '17, 01:28) $\log(n)^2$ for each query (30 Aug '17, 01:49) a_d_i2★ no.. while creating tree itself.. Can you post some pseudo code here..?? (30 Aug '17, 01:54) 2 @kauts_kanu, this is also known as mergesort tree. But I had to use input via getchar_unlocked() to get overcome TLE, so be careful about that @phantomhive. (30 Aug '17, 09:00) meooow ♦6★ Did not know about this.. thanks.. :) (30 Aug '17, 15:46)
 1 You can solve this problem by using Offline solution method and by creating Balance Binary Search Tree incrementally.. First iterate over all queries and store for each index what is query on that.. While creating Balanced BST you can find out number of smaller elements till now in log(N) using Binary Search tree that we have made till that point, and store answer for that pair(i,k) in some dict ansThatWeFound. Now iterate through query again, for ans(i,j,k) = ansThatWeFound(j,k) - ansThatWeFound(i-1,k). Hope this was clear enough and was providing enough info to approach this question now.. :) I you have some doubt then you can ask that in comments.. answered 30 Aug '17, 01:24 1.1k●1●10 accept rate: 19% can i do this with fenwick trees ?? amm creating the tree with sorted subarray will cost me N log(n) log(logn) and for each query it will cost log(n)*log(log(n)) (30 Aug '17, 22:28)
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question asked: 30 Aug '17, 01:04

question was seen: 524 times

last updated: 05 Feb '18, 20:23