@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)
>
>
> >
>
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