I think the greedy method of taking the current minimum sized 2 ropes and tying them will do. Consider this algo:- int getMinCost(){
priority_queue pq; insert all thread sizes to pq; int sum=0; while(!pq.empty()){ int a=pq.extractmin(); //O(logn) int b=pq.extractmin(); sum+=a+b; pq.push_back(a+b); } return sum; } Time: O(nlogn) Let me know if this fails in some case. Anurag On Mon, Mar 28, 2011 at 12:11 PM, bittu <shashank7andr...@gmail.com> wrote: > you are given n ropes,maybe of different length. the cost of tying two > ropes is the sum of their lengths.Find a way to tie these ropes > together so that the cost is minimum. > > > > Thanks > Shashank > > -- > 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.