I got the correct answer: If it is a valid array, sum of all elements in the array = value calculated using arithmetic progression formula
In this case Sum of arithmetic progression = (n/2)[2*a+(n-1)*d} where a = min of the array n = number of elements d = 1 If this value is equal to sum of all the elements in the array then it is a valid array On Dec 10, 6:44 am, ADITYA KUMAR <aditya...@gmail.com> wrote: > @jai > yeah, it can be done using count sort logic > but that will take O(n) extra space > > which can be avoided by using XOR. > > > > > > On Fri, Dec 10, 2010 at 3:34 AM, jai gupta <sayhelloto...@gmail.com> wrote: > > Algo: > > In first traverse find the min and the max values. > > if (max-min) not equals (N-1) > > return false > > In next traverse map each in a hashtable of size N where index=key-min. Now > > in case of collision return false > > return true > > > -- > > You received this message because you are subscribed to the Google Groups > > "Algorithm Geeks" group. > > To post to this group, send email to algoge...@googlegroups.com. > > To unsubscribe from this group, send email to > > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > > . > > For more options, visit this group at > >http://groups.google.com/group/algogeeks?hl=en. > > -- > Regards > Aditya Kumar > B-tech 3rd year > Computer Science & Engg. > MNNIT, Allahabad.- Hide quoted text - > > - Show quoted text - -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@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.
