@neeraj ur worst case complexity will be O(mn)
On 10/8/10, snehal jain <[email protected]> wrote: > > @tech > the ouput will be abhgrtsghgrthswert as no suffix of 1st matches with > prefix of 2nd > > > On 10/7/10, ligerdave <[email protected]> wrote: >> >> use pointer. shift to left if one more leading char has been found. >> any unmatched char resets the pointer to first char >> >> once you went through the entire list(first one), the pointer on the >> second list tells you where to concatenate >> >> that gives you O(n) where n is the length of first list >> >> On Oct 7, 3:52 am, snehal jain <[email protected]> wrote: >> > There are two linked list, both containing a character in each node. >> > >> > If one linked list contain characters o x e n c and second contain >> > characters e n c a r t a then the final linked list should contain o x >> > e n c a r t a i.e. if the end of one list is same as the start of >> > second then those characters should come only once. >> > >> > can we do it in O(n+m) where n and m are the length of list. both are >> > singly link list. >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to [email protected]. >> To unsubscribe from this group, send email to >> [email protected]<algogeeks%[email protected]> >> . >> 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 [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
