FineLine's blog

HELP REQUIRED IN SPOJ K-QUERY
By FineLine, history, 8 years ago, In English

Hi everyone,I have been trying this question SPOJ K-QUERYfor 2 days now.After unsuccessful attempts to come up with logic,I finally referred Blogs available online.But I am neither getting head nor tail of the logic. Can someone please tell me the formation of fenwick/segment tree that is used(especially the technique used).It will be a great help.Please! Thanks in advance.

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8 years ago, # |
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try amd blog on segment tree he has explained it nicely segment tree. If you still need help then ask.

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8 years ago, # |
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That has been explained so many times... Just google it.

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    8 years ago, # ^ |
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    I had tried it.But I did not get good explanation .All the blogs gave code .I tried understanding it using pen and paper literally.But I could not get the basic strategy behind it.The best blog I got was of PrinceOfPersia , But still could not understand it. Instead of showing me How to Google ,It would have been great that you could have explained it to me. Anyways Thanks radoslav11....

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      8 years ago, # ^ |
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      Well the main idea is to solve the problem offline. At every query we should care just for elements with value < K. Lets make events. Every event will be either an adding of element at position POS with value VAL or a query.

      We sort these events by VAL and K (depends if event is an adding or a query). Then our problem becomes:

      For every query what is the number of added elements before it in range [L; R].

      This can be done with an range-query and point update on segment tree or with a BIT: For every adding we set the value of POS with 1 and for every query we get sum of [L; R].

      The complexity is O(Nlog(N) + (M+N)*log(N)) = O(Nlog(N)).

      PS: The problem can be also solved online with a merge sort tree in O(N*log^2(N)) time.

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        8 years ago, # ^ |
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        Thanks radoslav11 and SORRY if you found my above comment offensive .I actually had tried lot before asking on the forum.Again Sorry!!

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