kumar wrote: > kerry wrote: > > > need O(n) time > > the array is set[n], > > for(i = 1, i <=n ; i++) > > if((set[i] & 1) & (set[i + 1] & 1)) > > then miss the number i+1; > > ur logic is not satisfied for even the array is in sorted order also. > For sorted array > we need ( set [ i ] & 1) & ( set [ i + 1 ] & 1) | | ( set [ i ] & 0 ) > & ( set [ i + 1 ] & 0 ) > then only ur logic is applicable > > Check it once. > If it satisfies give one example to understand code clearly.
even now, it won't work if two consecutive numbers are skipped. By the way: you can 'sort' the numbers into an array if you have lots of memory and the array subscription is cheap to do, just put number i into ith position. use the algorithms mentioned before in this thread --~--~---------~--~----~------------~-------~--~----~ 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 [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
