@ *$* : would you elaborate more on the problem ?
Because when the array will be simply sorted, even then the problem looks to
be solved.
I think we required something else that is not clear from the statement.
*Regards
Sanju
Happy to Help :)*
On Fri, Aug 19, 2011 at 2:10 AM, Abhishek Yadav <abhishek30.nit...@gmail.com
> wrote:
> Its the same as we do merge sort where we merge the two sorted array into
> one which will require an extra array......
> Is there any algorithm for inplace mergesort...?
>
> On Fri, Aug 19, 2011 at 2:09 PM, sagar pareek <sagarpar...@gmail.com>wrote:
>
>> Can be done in O(n) time but it will need O(n) space too
>>
>> take another array of same length
>>
>> then its code will be
>>
>> for( i=0,j=0,k=n/2+1 ;i<=n/2&&k<n; )
>> {
>> if(arr[i]>arr[k])
>> new[j++]=arr[k++];
>> else
>> new[j++]=arr[i++];
>> }
>>
>> if(k<n)
>> {
>> while(i<=n/2)
>> new[j++]=arr[i++]
>> }
>> else
>> {
>> while(j<n)
>> new[j++]=arr[k++]
>>
>> }
>>
>> On Fri, Aug 19, 2011 at 12:40 AM, *$* <gopi.komand...@gmail.com> wrote:
>>
>>> Sort an array of n positive integers containing n/2 sorted integers in
>>> first and second-half?
>>> in O(n) time complexity ..
>>> and space complexity should be constant
>>>
>>>
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>>
>>
>>
>> --
>> **Regards
>> SAGAR PAREEK
>> COMPUTER SCIENCE AND ENGINEERING
>> NIT ALLAHABAD
>>
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>
>
>
> --
> Abhishek Yadav
> Comp Engg.
> NIT Kurukshetra
>
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