Using segment tree: it does using O(logn) complexity just check it out: http://codepad.org/5jVxLmsA
On Sat, Jun 12, 2010 at 2:10 AM, Modeling Expert < cs.modelingexp...@gmail.com> wrote: > My previous post went unfinished :(( > anyway this is summary of algo . (as N is very large , sorting could > be costly so this method doesn't do that ) > > algo :: > have a map<int,bool> > > 1. For given number , if its less than Sum S , make map[S-number] = > true only if map[number] does not exist > 2. for Next number , if map[number] is true , we got a pair > ( number , map[number]) else do 1 > > For exampe S = 100 , if we get 20 , let's make map[80] = true ; > next if we get 80 , we will first check map[80] and if its true , we > get a pair. > > If there are repetations , we can take map of <int,int> where second > argument is count of first element , > say of 20 comes 4 times we will store map[20] = 4 > > > > > On Jun 12, 11:40 am, Chakravarthi Muppalla <chakri...@gmail.com> > wrote: > > I would use an array. > > > > a[] = 1 3 7 8 9 78 67 23 > > a[] = 1 3 7 8 9 23 67 78 //sort the array > > reqSum = 30; > > for i :a.length-1; i>=0; i-- > > t = reqSum - a[i]; > > if(t is negative) continue; > > find(t);//in the rest of the array;(binary search) > > if(t found in the array) return index of t, current value of i; > > I guess it works.(we may have to use a hash map to speed it up in the > long > > run). > > > > On Sat, Jun 12, 2010 at 10:29 AM, Rohit Saraf > > <rohit.kumar.sa...@gmail.com>wrote: > > > > > > > > > I guess it can be done in efficiently using a simple divide and conquer > > > scheme almost imitating mergesort. > > > Can you think of it now? :D > > > > > -------------------------------------------------- > > > Rohit Saraf > > > Second Year Undergraduate, > > > Dept. of Computer Science and Engineering > > > IIT Bombay > > >http://www.cse.iitb.ac.in/~rohitfeb14 > > > > > On Fri, Jun 11, 2010 at 10:07 PM, sharad kumar < > sharad20073...@gmail.com>wrote: > > > > >> all possible pairs > > > > >> -- > > >> You received this message because you are subscribed to the Google > Groups > > >> "Algorithm Geeks" group. > > >> To post to this group, send email to algoge...@googlegroups.com. > > >> To unsubscribe from this group, send email to > > >> algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > <algogeeks%2bunsubscr...@googlegroups.com<algogeeks%252bunsubscr...@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 algoge...@googlegroups.com. > > > To unsubscribe from this group, send email to > > > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > <algogeeks%2bunsubscr...@googlegroups.com<algogeeks%252bunsubscr...@googlegroups.com> > > > > > . > > > For more options, visit this group at > > >http://groups.google.com/group/algogeeks?hl=en. > > > > -- > > Thanks, > > Chakravarthi. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algoge...@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@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 algoge...@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.
