No. At step 6, we have found a 1 in an even spot and a 0 in an odd
spot, so we switch them. Switching a 1 and a 0 simply means storing 0
where the 1 was and storing 1 where the 0 was; there is no need to use
the usual code to interchange two values.

Dave

On Nov 13, 8:37 am, "Vikram Venkatesan" <[EMAIL PROTECTED]>
wrote:
> @Dave,
>  >>*Otherwise, set the even position to 0 and the odd position to 1.*
>     I think your solution might be inserting 0's and 1's into the array from
> nowhere (thus filling the whole array with alternating 0's and 1's up to the
> given size !). The question is to re-arrange existing elements in the array.
>
> -Vikram
>
> On 11/13/07, Dave <[EMAIL PROTECTED]> wrote:
>
>
>
>
>
> > The following algorithm examine the contents of each element of the
> > array at most once.
>
> > 1. Start with the first even position and the first odd position.
> > 2. Search forward through the even positions until you reach the end
> > of the array or find a 1, whichever comes first.
> > 3. Search forward through the odd positions until you reach the end of
> > the array or find a 0, whichever comes first.
> > 4. If you reached the end of the array in either of the above
> > searches, you are finished.
> > 5. Otherwise, set the even position to 0 and the odd position to 1.
> > 6. Repeat steps 2-6.
>
> > Dave
>
> > On Nov 13, 7:39 am, geekko <[EMAIL PROTECTED]> wrote:
> > > you are given an array of integers containing only 0s and 1s. You have
> > > to place all the 0s in even positions and 1s in odd position. And if
> > > suppose, no. of 0s exceeds no. of 1s or vice versa keep them
> > > untouched. Do in ONE PASS without taking extra memory.(modify array in
> > > place)- Hide quoted text -
>
> - Show quoted text -


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