Check this out, it works in O(n);
int i = 0;
int j = n-1;
while((i<n && j>= 0)&&(i<j))
{
if(a[i]>0 && a[j]<0)
{
swap(a[i],a[j]);
i++; j--;
}
else
{
if(a[i]<0)
i++;
if(a[j]>0)
j--;
}
}
Thanks,
Mani
On Thu, Jun 14, 2012 at 1:05 PM, Mad Coder <[email protected]> wrote:
>
> As nothing is written about space complexity, I am assuming that we can
> take extra space.
>
> Take a temporary array of length n.
>
> 1. Maintain a counter for the length of temporary array filled till now.
>
> 2. Traverse the given array. If value contained is negative insert it in
> new array and update counter. After complete traversal all negative values
> will be in the temporary array.
>
> 3. Traverse again the given array. Repeat step 2 but this time for
> positive numbers.
>
> Finally temporary array contains the required answer. If required copy it
> into original array.
>
> As this approach takes max. 3 traversals so its complexity is O(n).
>
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Thanks & Regards,
Mani
http://www.sanidapa.com - The music Search engine
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