how to do it in space comp- O(1)  . I mean without using additional arrays.
Could it be done ?

On Thu, Jun 14, 2012 at 3:46 PM, utsav sharma <utsav.sharm...@gmail.com>wrote:

> @all :- by segregating 0 and 1 or by taking quicksort approach
> we have to swap no.s resulting which array becomes unordered.
>
> my approach is start from right and  if a negative no. occurs insert it
> into another array temp[]
> this will shift all positive no.s to right and in 2nd pass start from left
> of original array and insert all the elements of temp.
> time comp-O(n), space comp-O(n)
> correct me if i am wrong
>
> On Thu, Jun 14, 2012 at 1:53 PM, saurabh singh <saurab...@gmail.com>wrote:
>
>> 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 <
>> manikantabab...@gmail.com> 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 <imamadco...@gmail.com>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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>
>
>
> --
> Utsav Sharma,
> NIT Allahabad
>
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