WE CAN USE RADIX SORT TO FIND THE MAX VERSION :TRAVERSING FROM LEFT TO RIGHT
(RADIX WILL BE NUMBERS BETWEEN THE POINTS).
It will reduce the search space too.
On Tue, Oct 11, 2011 at 1:58 AM, sumit kumar pathak
<sumitkp1...@gmail.com>wrote:
> *
> *regards
> - Sumit Kumar Pathak
> (Sumit/ Pathak/ SKP ...)
> *Smile is only good contagious thing.*
> *Spread it*!!!!!
>
>
>
> On Tue, Oct 11, 2011 at 11:22 AM, DIPANKAR DUTTA <
> dutta.dipanka...@gmail.com> wrote:
>
>> what's happen if the versions are not in same length..
>>
>> For example
>>
>> v1: 1.1.1.133.2
>> v2: 1.2
>> v3: 1.2.3.4.5555.333
>> v4: 1.2.3.4.5554.222
>> v5: 1.3.2.2.2.2.2.2.2.2.2.2.2.2
>>
>> It implies we must be scan from left side one by one ...
>>
>> In general the we make a some sort of lexical comparison where each
>> character is a number and separated by number ..
>>
>> the problem definition is as :
>>
>> let V1,V2,V3 ...Vn be the n version
>>
>> let S={1,2,3...n} ,index=0
>>
>> Latest[S,index]= return Vi if { Max(ALL Vi[k] where i belongs to S )}
>> is a singleton, else
>> = return Latest[ set of all index belongs to { Max(ALL
>> Vi[k] where i belongs to S )}, index+1 ]
>>
>>
>>
>>
>>
>> On 10/11/11, Dave <dave_and_da...@juno.com> wrote:
>> > @Karen: It is more complicated than scanning character by character.
>> > E.g., "1.10.3" is older than "1.9.7". I think
>>
>
> <snip>
>
>> you need to parse the
>> > numbers between the dots and compare them
>>
> </snip>
>
> simple, easier and correct way.
> points:
> - compare left to right (ofcourse) each version being vector of numbers
> - for diffrent size, assume extra ones to be zero while comapring (no
> need to store trailing zeros)
>
>
>
>> one by one. Thus, in the
>> > above example, 1 compares equal to 1, so you keep scanning. Then 10
>> > compares greater than 9 so the first string is number of the newer
>> > version. I did this many years ago in a csh install script for a unix
>> > product.
>> >
>> > Dave
>> >
>> > On Oct 10, 9:52 pm, "bagaria.ka...@gmail.com"
>> > <bagaria.ka...@gmail.com> wrote:
>> >> Given two strings describing the version of a particular software need
>> to
>> >> find the later version.
>> >>
>> >> For eg.
>> >> 1st string = "1.2.4.5"
>> >> 2nd string="1.2.3.5"
>> >>
>> >> 1st string is the later one.
>> >>
>> >> Can be done using traversing the string and comparing each character
>> one
>> >> after the another. Looking for a better solution with lesser
>> complexity.
>> >>
>> >> --
>> >> Thanks and Regards
>> >>
>> >> *Karan Bagaria*
>> >> *MCA Final Year*
>> >> Training and Placement Representative
>> >> *NIT Durgapur*
>> >
>> > --
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>> >
>>
>>
>> --
>> Thanks and Regards,
>> ------------------------------
>> **DIPANKAR DUTTA
>> Software Development Engineer
>> Xen Server - OpenStack Development Team (DataCenter and Cloud)
>>
>> Citrix R&D India Pvt Ltd
>> 69/3, Millers Road, Bangalore – 560052
>> Phone: +91 8147830733
>> Office: Extn: 16429
>> Email: dipankar.du...@citrix.com
>>
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