Okay, forgetting the information that two parts are sorted and treating it
as any other array, we can sort in O(nlogn) using, say, heapsort. Is O(n)
possible with constant space ?

Thanks,
Balaji.

On Tue, Apr 12, 2011 at 9:25 PM, Balaji Ramani <[email protected]>wrote:

> But Vaibhav's solution I think is O(n^2). For example, when input is
>
> 101 102 103 104 1 2 3 4
>
> We first swap 1 and 101 and then bubble 101 to the end of the subarray 2 3
> 4 .
>
> This bubbling we must repeat after each swap.
>
> This results in n/2 + n/2-1 + n/2-2 + .. comparisons, which is O(n^2).
>
> Please correct me if I am wrong.
>
> Can this be solved in better than O(n^2) with constant space ?
>
> Thanks,
> Balaji.
>
>
> On Tue, Apr 12, 2011 at 8:43 PM, Carl Barton 
> <[email protected]>wrote:
>
>> That's linear space, not constant space.
>> Vaibhav's seems good for constant space solution
>>
>>
>> On 12 April 2011 13:17, sravanreddy001 <[email protected]> wrote:
>>
>>> Yes.. merge sort.
>>>
>>> O(n) to find the starting of 2nd sub-array.
>>> and O(n) for the merge process (similar to last step in merge sort)
>>>
>>> O(n)
>>>
>>> On Apr 12, 2:37 pm, Akash Agrawal <[email protected]> wrote:
>>> > Given an array with two subparts sorted. How will you make a final
>>> sorted
>>> > array.
>>> >
>>> > i/p:  1, 5, 7, 9, 11, 23, 2, 3, 8, 9, 21
>>> >
>>> > o/p:
>>> > 1, 2, 3, 5, 7, 8, 9, 9, 11, 21, 23
>>> >
>>> > Regards,
>>> > Akash Agrawalhttp://tech-queries.blogspot.com/
>>>
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