@Nishan :
For the given example, the number of elements which are not in order may be
less ..can u prove this ?
And also, how can u place an incorrect positioned number at correct
position ---> takes O(n) for each number ....
On Sun, May 26, 2013 at 8:55 PM, Ankit Agarwal <ankuagarw...@gmail.com>wrote:
> The first pass is not necessary. We can finding the middle element as
> follows:
>
> N = even, Range [ 0 - (N/2 - 1) ] & [ N/2 - (N - 1) ]
>
> N = odd,
> if (A[N/2] > A[N/2 -1]) Range [ 0 - N/2 ] & [ (N/2 + 1) -
> (N - 1) ]
>
> else if ( A[N/2] < A[N/2 + 1]) Range [ 0 - (N/2 - 1) ] & [
> N/2 - (N - 1) ]
>
>
>
>
>
> On Sun, May 26, 2013 at 8:28 PM, Nishant Pandey <
> nishant.bits.me...@gmail.com> wrote:
>
>> The solution could be given in this way.
>>
>> 1) In one pass get the end index of both array says e1 and e2.
>> 2) now in next pass compare elements at e1 and e2 .
>> a) if a(e1) > a(e2) swap the elements and then decreament e1 and e2
>> both.
>> b) if a(e1) < a(e2) decreament e2.
>> c) if a(e1) == a(e2) then swap a(e1) with a(e2-1) and then decrement
>> e1 by1 and e2 by 2.
>>
>> After this pass there may be one or two element not at coret position, so
>> their position can be placed just by shifting in elements in another pass.
>>
>> So as a total it would be O(n) but it requires 3 passes.
>>
>>
>> If some one is having something better tan this, please suggest.
>>
>>
>>
>> On Sun, May 26, 2013 at 6:46 PM, bharat b
>> <bagana.bharatku...@gmail.com>wrote:
>>
>>> An array is given, first and second half are sorted .. Make the array
>>> sorted inplace... Need an algo better than O(n^2)..
>>> If the length of the array is odd.. middle is either in first half or
>>> second half.
>>> Ex:
>>> 1. Arr[] = {2,3,6,8,-5,-2,3,8} --> output : Arr[]={-5,-2,2,3,3,6,8,8};
>>> 2. Arr[] = {2,3,6,8,-5,-2,3} --> output : Arr[]={-5,-2,2,3,3,6,8};
>>> 3. Arr[] ={2,3,6,-5,-2,3,8} --> output : Arr[]={-5,-2,2,3,3,6,8};
>>>
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>
>
>
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>
> *Ankit Agarwal*
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