@Rais: I don't see that it leaves the array unchanged if the number of
zeros does not equal the number of ones.
Dave
On Aug 19, 7:23 am, Rais Khan <raiskhan.i...@gmail.com> wrote:
> void ArrayShifting(int *str, int size)
> {
> int odd=1, even=0;
> while(odd < size)
> {
> int position;
> if(str[even]!=0)
> {
> position = even+1;
> while(position < size)
> {
> if(str[position]==0)
> { str[position]=1;str[even]=0;break;}
> position = position+2;
> }
> }
> even = even+2;
> if(str[odd]!=1)
> {
> position = odd+1;
> while(position < size)
> {
> if(str[k]==0)
> { str[position]=0;str[odd]=1;break;}
> position = position+2;
> }
> }
> odd=odd+2;
> }
>
> }
>
> This code seems working for me, If you agree then we can work on measuring
> the complexity & improving the code accordingly.
>
> On Thu, Aug 19, 2010 at 5:27 PM, Ashutosh Tamhankar
> <asshuto...@gmail.com>wrote:
>
>
>
> > Hi Amit
>
> > Am I allowed to keep counters to count the number of 1's and 0s right?
>
> > The condition of not using extra memory is for the array!?
>
> > Regards
> > Ashutosh
>
> > 2010/8/18 amit <amitjaspal...@gmail.com>
>
> > you are given an array of integers containing only 0s and 1s. you have
> >> to place all the 0s in even position and 1s in odd position and if
> >> suppose no if 0s exceed no. of 1s or vice versa then keep them
> >> untouched. Do that in ONE PASS and WITHOUT taking EXTRA MEMORY.
>
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