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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