@ gurav vivek

i think u r only exchanging odd and even.. only putting even to end and odd
in front and not doing descending sorting on odds and ascending on even..

plz correct me if i hv missed something..

On 24 June 2010 22:49, Vivek Sundararajan <[email protected]> wrote:

> Hi, how about this -
>
> Do a merge sort, now, while merging two sorted list, give more priority to
> odd numbers :)
>
> I believe this falls into the right solutions :)
>
> Any breaking cases?
>
>
> On 24 June 2010 09:41, Gaurav Singh <[email protected]> wrote:
>
>> I think in this case, bubble sorting will be a better idea. just
>> replace the condition of comparison with the condition that earlier
>> number is even and later number is odd. I mean we can do sumthing lyk
>> this :
>>
>> for i=1 to n-1
>>      for j=1 to n-i-1
>>             if iseven(ar[j]) AND (NOT iseven(ar[j+1]))
>>             then       swap both of them.
>>
>> Please correct me if I am wrong somewhere.
>>
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>
>
> --
> "Reduce, Reuse and Recycle"
> Regards,
> Vivek.S
>
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