@deepika anand
solution given by me is for getting number of swap's in O(n)
as far as sorting goes any O(n lgn) algo can be used .
if count sort is allowed then O(n) for sorting also.[constant extra space.. ]
On Sat, Jun 23, 2012 at 12:49 AM, ashish jain <ashishjainco...@gmail.com> wrote:
> @all
>
> yaa.. For getting number of swaps.. we have to calculate total number of
> zeroes on the right for each '1' and on adding them
> we will get the number of swaps. and in O(n) time.
>
>
> On Sat, Jun 23, 2012 at 1:16 PM, Guruprasad Sridharan
> <sridharan.mi...@gmail.com> wrote:
>>
>> It will work because we need to swap only adjacent elements. So we do a
>> sweep from left to right finding the number of ones encountered to the left
>> of a zero when we find a zero. We keep adding the number as we sweep the
>> entire length.
>>
>>
>> On Sat, Jun 23, 2012 at 1:14 PM, deepikaanand <swinyanand...@gmail.com>
>> wrote:
>>>
>>> @saurabh..wat array r u getting finally....is it all 1's or in sorted
>>> order for the input
>>> int a[8]={1,1,1,0,1,0,1,1};
>>>
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