Order may not be maintained necessarily by this solution
Saurabh Singh
B.Tech (Computer Science)
MNNIT
blog:geekinessthecoolway.blogspot.com



On Thu, Jun 14, 2012 at 1:47 PM, Manikanta Babu
<[email protected]>wrote:

> 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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