@hemesh,kk: let's take a test case : arr : 2 4 6 8 arr^2 : 6 8 10 10 12 14 (sum of each unique pair in arr[i])
let's say target sum is 26 your solution will return true as they 12+14=26 but in 12 and 14, 8 is common, infact 26 is not possible in the given array can u please elaborate how will you take care of such situation ? @jalaj: yes it's O( (n^3)*logn) @bhavesh: fyi.. log(n^3)=3*log(n)=O(log(n)) so it's same.. :P -- Amol Sharma Final Year Student Computer Science and Engineering MNNIT Allahabad <http://gplus.to/amolsharma99> <http://twitter.com/amolsharma99><http://in.linkedin.com/pub/amol-sharma/21/79b/507><http://www.simplyamol.blogspot.com/><http://facebook.com/amolsharma99> On Mon, Jun 18, 2012 at 12:29 AM, KK <kunalkapadi...@gmail.com> wrote: > @Hemesh : +1 > @Jalaj : read Hemesh's solution again it is for 4sum. > In short, make a new array having sum of each unique pair of given array. > -> O(n^2) > sort it -> O(n^2) > for each number bi in new array, binary search (target - bi) in the same > array -> O(n^2) > > > On Sunday, 17 June 2012 12:41:40 UTC+5:30, Jalaj wrote: >> >> The solution which hemesh gave was solution to 3SUM hard problem the best >> solution for which can be achieved in n^2 . >> And the original question is a kind of 4SUM hard problem for which best >> possible solution i think is again n^3 and Amol what you told is not n^3 , >> finding all triplets will itself take n^3 and doing a binary search again >> that sums upto n^3*logn. >> >> @shashank it is not a variation of 3SUM problem as in 3SUM problem a+b+c >> = some constant , but in your case it is "b+c+d = s-a", where a can change >> again and again so if you do even apply 3SUM logic to it you will have to >> do it for every a which will make it n^2*n = n^3 >> >> >> >> On Sat, Jun 16, 2012 at 2:45 AM, sanjay pandey <sanjaypandey...@gmail.com >> > wrote: >> >>> @hemesh cud u plz elaborate wat is b[k]=a[i]+a[j]...n also ur >>> solution... >>> >>> -- >>> 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+unsubscribe@** >>> googlegroups.com <algogeeks%2bunsubscr...@googlegroups.com>. >>> For more options, visit this group at http://groups.google.com/** >>> group/algogeeks?hl=en <http://groups.google.com/group/algogeeks?hl=en>. >>> >> >> >> >> >> -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To view this discussion on the web visit > https://groups.google.com/d/msg/algogeeks/-/9jCCN5iHDB8J. > > 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.
