@Anshu: My logic: during merge step, lets say you have two array left and right (both are sorted) and you are going to merge it.
l[i] represent the element of left arrray r[j] represent the element of right array if (r[j] < l[i]) { increase the inversion count by the number of elements lefts in left array put element r[j] into merged array j++ } else { no increase in count, just put the element r[i] into merged array i++; This will be O(nlogn) solution. Can we do better using BIT or segment trees? Can you please provide me some hint of how to solve it using segment trees, I just know the basics of it.. On Thu, May 12, 2011 at 5:00 PM, anshu mishra <anshumishra6...@gmail.com>wrote: > explain your logic instead of posting code, use binary index tree or > segment tree or bst to solve this problem. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.