I think i hve figured out the actual answer .
Suppose we maintain a queue of N words in the memory. With two things 1. front 2. rear As a new word enters (recognized by a space) -> front = front.next; if(it is already there in the list ) ++frequency of occurence; else{ -> temp = rear; -> rear = newWord; -> temp.next = rear; (simple insertion at the rear end of queue) } I think this will work .. if Anyone thinks it won't then Do suggest modification or indicate error Pawandeep --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---