check this out Let x and y be the missing number,
Now equation 1 is : x + y = [n(n+1)/2] - S equation 2 is: x * y = N! /P solve both we get elements On Fri, Jul 31, 2009 at 8:27 PM, Devi G <devs...@gmail.com> wrote: > The logic is actually simple. Tot if we mark in some way an element when > it's scanned, we can find the missing numbers in the second scannin. > > 3,5,1,2,9,10,8,6 > > When for loop sees '3' it knows elt 3 is there. So multiplies the number at > 3rd position by some arbitrary number. (* I've taken the arbitrary number > to be n here but CORRECT ONE IS n+3 cos n will fail in some cases*) > > so, when it sees '5' multiplies the number at 5th position by n+3. > It skips when the numbr is greater than n. > > n+3 = 11 here. > > So,after first loop, > 33, 55, 11, 2 , 99, 110, 8, 66. > > So now, in the second scan, the indices of all elts that are divisible by > n+3 are present in the array. > elts at 4th and 7th positions are not divisible. hence missing numbers are > 4 and 7. > > > > > > --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
