Re: [algogeeks] Re: Find the later version
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.comwrote:
*
*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..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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--
**DIPANKAR DUTTA
Software Development Engineer
Xen Server - OpenStack Development Team (DataCenter and Cloud)
Citrix RD India Pvt Ltd
69/3, Millers Road, Bangalore – 560052
Phone: +91 8147830733
Office: Extn: 16429
Email: dipankar.du...@citrix.com
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[algogeeks] Re: Find the later version
As others have said, you have not completely specified the problem.
If you always have exactly 4 base 10 numbers separated by dots, and
you are working in C or C++ then you can use sscanf to get the numbers
and compare them:
// Return:
// positive if version s1 is an earlier version than s2,
// negative if s2 is earlier than s1
// zero if they're the same version
#define VERSION_CMP_ERROR INT_MAX
int version_cmp(char *s1, char *s2)
{
int i, v1[4], v2[4];
if (sscanf(s1, %u.%u.%u.%u, v1+0, v1+1, v1+2, v1+3) != 4 ||
sscanf(s2, %u.%u.%u.%u, v2+0, v2+1, v2+2, v2+3) != 4)
return VERSION_CMP_ERROR;
for (i = 0; i 4; i++)
if (v1[i] != v2[i])
return v2[i] - v1[i];
return 0;
}
On Oct 11, 4:52 am, 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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Re: [algogeeks] Re: Find the later version
scan the numbers as string if both strings dont have equal numbers of dots then add trailing 0's with dots to the string with less number of dots now divide both the string into 2 equal halves compare the left half .if unequal then divide recursively and again compare left of the divided string.if equal the compare the right half and divide and compare it recursivelythis will reduce the complexity to logn. -- Amol Sharma Third Year Student Computer Science and Engineering MNNIT Allahabad http://gplus.to/amolsharma99 http://twitter.com/amolsharma99http://in.linkedin.com/pub/amol-sharma/21/79b/507http://youtube.com/amolsharma99 On Thu, Oct 13, 2011 at 2:21 AM, sravanreddy001 sravanreddy...@gmail.comwrote: it is expected that the version is given as string, in that case, atoi() has be used to convert string to int -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Find the later version
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..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 you need to parse the
numbers between the dots and compare them 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*
--
You received this message because you are subscribed to the Google Groups
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algogeeks+unsubscr...@googlegroups.com.
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--
Thanks and Regards,
--
**DIPANKAR DUTTA
Software Development Engineer
Xen Server - OpenStack Development Team (DataCenter and Cloud)
Citrix RD India Pvt Ltd
69/3, Millers Road, Bangalore – 560052
Phone: +91 8147830733
Office: Extn: 16429
Email: dipankar.du...@citrix.com
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Re: [algogeeks] Re: Find the later version
*
*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..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*
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
--
Thanks and Regards,
--
**DIPANKAR DUTTA
Software Development Engineer
Xen Server - OpenStack Development Team (DataCenter and Cloud)
Citrix RD India Pvt Ltd
69/3, Millers Road, Bangalore – 560052
Phone: +91 8147830733
Office: Extn: 16429
Email: dipankar.du...@citrix.com
--
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[algogeeks] Re: Find the later version
@Karen: It is more complicated than scanning character by character. E.g., 1.10.3 is older than 1.9.7. I think you need to parse the numbers between the dots and compare them 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* -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
