I think he means to edit the comparison function to get the order. so
at a time only 2 elements are compared.

On May 28, 7:51 am, Logic King <crazy.logic.k...@gmail.com> wrote:
> @sunny it will work fine if you have 2 numbers only...but what about the
> list...3..4 or 5..or more....then the possible number of combinations will
> be  'N!'...where n is the number of digits.......the code will work quite
> slowly for larger 'n'.
>
>
>
>
>
>
>
> On Fri, May 27, 2011 at 3:33 PM, Dave <dave_and_da...@juno.com> wrote:
> > @Shubham: Try 8, 89, 7. Your algorithm gives 8897, but the right
> > answer is 8987.
>
> > Dave
>
> > On May 27, 1:11 pm, shubham <shubh2...@gmail.com> wrote:
> > > check whether these steps work:
>
> > > step 1:
> > >         sort the given numbers in the decreasing order based on their
> > first
> > > digit.
> > > step 2:
> > >         if two numbers come out to be equal in the above case & both of
> > > their next digit exist then sort on the basis of their next digit,
> > otherwise
> > > the number
> > >         whose next digit doesnot exist should be placed before the other
> > > number.
> > > step 3:
> > >        concatenate these numbers.
>
> > > e.g.
>
> > > (0,1,10,100) sorting it gives: 1,10,100,0 => 1101000
> > > (97,8,9) sorting gives: 9,97,8 => 9978
>
> > > correct me if i'm wrong.
>
> > > Shubham
>
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