Re: [algogeeks] Regex tester
ya, it is correct, i misunderstood it..
any optimization on the same though ?
On Fri, Dec 28, 2012 at 9:55 AM, shady wrote:
> @ritesh
> umm, well here's a simple testcase to show the problem in the code..
> isMatch("aa", "a*")
>
>
> On Thu, Dec 27, 2012 at 7:17 PM, Ritesh Mishra wrote:
>
>> @shady : either the string will be stored in heap or stack. thus
>> accessing address in heap or stack is not going to give u seg fault . and
>> rest things are very well handled in the code :)
>> As saurabh sir has explained in thread
>> https://mail.google.com/mail/u/1/#inbox/13ba918bdb9aac9e
>> when seg fault occurs .
>> Regards,
>>
>> Ritesh Kumar Mishra
>> Information Technology
>> Third Year Undergraduate
>> MNNIT Allahabad
>>
>>
>> On Thu, Dec 27, 2012 at 6:43 PM, ~*~VICKY~*~ wrote:
>>
>>> I'm giving you a simple recursive code which i wrote long back. Please
>>> let me know if it fails for any cases. Ignore the funny cout's It used to
>>> help me debug and i'm lazy to remove it. :P :)
>>>
>>> #include
>>> #include
>>> using namespace std;
>>> /*
>>> abasjc a*c
>>> while(pattern[j] == '*' text[i] == pattern[j]) {i++; j++}
>>> */
>>> bool match(string text, string pattern, int x, int y)
>>> {
>>> if(pattern.length() == y)
>>> {
>>> cout<<"hey\n";
>>> return 1;
>>> }
>>> if(text.length() == x)
>>> {
>>> cout<<"shit\n";
>>> return 0;
>>> }
>>> if(pattern[y] == '.' || text[x] == pattern[y])
>>> {
>>> cout<<"in match"<>> return match(text,pattern,x+1,y+1);
>>> }
>>> if(pattern[y] == '*')
>>> return match(text,pattern,x+1,y) || match(text,pattern,x+1,y+1)
>>> || match(text,pattern,x,y+1);
>>>
>>> if(text[x] != pattern[y])
>>> {
>>> cout<<"shit1\n";
>>> return 0;
>>> }
>>>
>>> }
>>>
>>> int main()
>>> {
>>> string text,pattern;
>>> cin >> text >> pattern;
>>> cout << match(text, pattern,0, 0);
>>> }
>>>
>>> On Thu, Dec 27, 2012 at 6:10 PM, shady wrote:
>>>
Thanks for the link Ritesh,
if (isMatch(s, p+2)) return true;
isnt this line incorrect in the code, as it can lead to segmentation
fault... how can we directly access p+2 element, we know for sure that p is
not '\0', but p+1 element can be '\0' , therefore leading to p+2 to be
undefined.
On Thu, Dec 27, 2012 at 6:23 AM, Ritesh Mishra wrote:
> try to solve it by recursion ..
> http://www.leetcode.com/2011/09/regular-expression-matching.html
>
>
> Regards,
>
> Ritesh Kumar Mishra
> Information Technology
> Third Year Undergraduate
> MNNIT Allahabad
>
>
> On Sun, Dec 23, 2012 at 11:14 PM, Prem Krishna Chettri <
> hprem...@gmail.com> wrote:
>
>> Well I can tell you Something about design pattern to solve this
>> case..
>>
>>What I mean is by using The State Machine Design Pattern,
>> Anyone can solve this. but Ofcourse it is complicated.
>>
>>
>>
>>
>> On Sun, Dec 23, 2012 at 11:01 PM, shady wrote:
>>
>>> that's the point, Have to implement it from scratch... otherwise
>>> java has regex and matcher, pattern to solve it...
>>>
>>>
>>> On Sun, Dec 23, 2012 at 10:28 PM, saurabh singh >> > wrote:
>>>
If you need to implement this for some project then python and java
have a very nice library
Saurabh Singh
B.Tech (Computer Science)
MNNIT
blog:geekinessthecoolway.blogspot.com
On Sun, Dec 23, 2012 at 7:48 PM, shady wrote:
>
> http://stackoverflow.com/questions/13144590/to-check-if-two-strings-match-with-alphabets-digits-and-special-characters
>
> any solution for this. we need to implement such regex
> tester
>
> some complex cases :
> *string** regex * -> * status*
> *
> *
> reesd re*.d -> match
> re*eed reeed -> match
>
> can some one help with this ?
>
> --
>
>
>
--
>>>
>>> --
>>>
>>>
>>>
>>
>> --
>>
>>
>>
>
> --
>
>
>
--
>>>
>>>
>>>
>>> --
>>> Cheers,
>>>
>>> Vicky
>>>
>>> --
>>>
>>>
>>>
>>
>> --
>>
>>
>>
>
>
--
Re: [algogeeks] Regex tester
@ritesh
umm, well here's a simple testcase to show the problem in the code..
isMatch("aa", "a*")
On Thu, Dec 27, 2012 at 7:17 PM, Ritesh Mishra wrote:
> @shady : either the string will be stored in heap or stack. thus accessing
> address in heap or stack is not going to give u seg fault . and rest things
> are very well handled in the code :)
> As saurabh sir has explained in thread
> https://mail.google.com/mail/u/1/#inbox/13ba918bdb9aac9e
> when seg fault occurs .
> Regards,
>
> Ritesh Kumar Mishra
> Information Technology
> Third Year Undergraduate
> MNNIT Allahabad
>
>
> On Thu, Dec 27, 2012 at 6:43 PM, ~*~VICKY~*~ wrote:
>
>> I'm giving you a simple recursive code which i wrote long back. Please
>> let me know if it fails for any cases. Ignore the funny cout's It used to
>> help me debug and i'm lazy to remove it. :P :)
>>
>> #include
>> #include
>> using namespace std;
>> /*
>> abasjc a*c
>> while(pattern[j] == '*' text[i] == pattern[j]) {i++; j++}
>> */
>> bool match(string text, string pattern, int x, int y)
>> {
>> if(pattern.length() == y)
>> {
>> cout<<"hey\n";
>> return 1;
>> }
>> if(text.length() == x)
>> {
>> cout<<"shit\n";
>> return 0;
>> }
>> if(pattern[y] == '.' || text[x] == pattern[y])
>> {
>> cout<<"in match"<> return match(text,pattern,x+1,y+1);
>> }
>> if(pattern[y] == '*')
>> return match(text,pattern,x+1,y) || match(text,pattern,x+1,y+1)
>> || match(text,pattern,x,y+1);
>>
>> if(text[x] != pattern[y])
>> {
>> cout<<"shit1\n";
>> return 0;
>> }
>>
>> }
>>
>> int main()
>> {
>> string text,pattern;
>> cin >> text >> pattern;
>> cout << match(text, pattern,0, 0);
>> }
>>
>> On Thu, Dec 27, 2012 at 6:10 PM, shady wrote:
>>
>>> Thanks for the link Ritesh,
>>> if (isMatch(s, p+2)) return true;
>>> isnt this line incorrect in the code, as it can lead to segmentation
>>> fault... how can we directly access p+2 element, we know for sure that p is
>>> not '\0', but p+1 element can be '\0' , therefore leading to p+2 to be
>>> undefined.
>>>
>>> On Thu, Dec 27, 2012 at 6:23 AM, Ritesh Mishra wrote:
>>>
try to solve it by recursion ..
http://www.leetcode.com/2011/09/regular-expression-matching.html
Regards,
Ritesh Kumar Mishra
Information Technology
Third Year Undergraduate
MNNIT Allahabad
On Sun, Dec 23, 2012 at 11:14 PM, Prem Krishna Chettri <
hprem...@gmail.com> wrote:
> Well I can tell you Something about design pattern to solve this
> case..
>
>What I mean is by using The State Machine Design Pattern,
> Anyone can solve this. but Ofcourse it is complicated.
>
>
>
>
> On Sun, Dec 23, 2012 at 11:01 PM, shady wrote:
>
>> that's the point, Have to implement it from scratch... otherwise java
>> has regex and matcher, pattern to solve it...
>>
>>
>> On Sun, Dec 23, 2012 at 10:28 PM, saurabh singh
>> wrote:
>>
>>> If you need to implement this for some project then python and java
>>> have a very nice library
>>>
>>>
>>> Saurabh Singh
>>> B.Tech (Computer Science)
>>> MNNIT
>>> blog:geekinessthecoolway.blogspot.com
>>>
>>>
>>> On Sun, Dec 23, 2012 at 7:48 PM, shady wrote:
>>>
http://stackoverflow.com/questions/13144590/to-check-if-two-strings-match-with-alphabets-digits-and-special-characters
any solution for this. we need to implement such regex
tester
some complex cases :
*string** regex * -> * status*
*
*
reesd re*.d -> match
re*eed reeed -> match
can some one help with this ?
--
>>>
>>> --
>>>
>>>
>>>
>>
>> --
>>
>>
>>
>
> --
>
>
>
--
>>>
>>> --
>>>
>>>
>>>
>>
>>
>>
>> --
>> Cheers,
>>
>> Vicky
>>
>> --
>>
>>
>>
>
> --
>
>
>
--
Re: [algogeeks] Regex tester
@shady : either the string will be stored in heap or stack. thus accessing
address in heap or stack is not going to give u seg fault . and rest things
are very well handled in the code :)
As saurabh sir has explained in thread
https://mail.google.com/mail/u/1/#inbox/13ba918bdb9aac9e
when seg fault occurs .
Regards,
Ritesh Kumar Mishra
Information Technology
Third Year Undergraduate
MNNIT Allahabad
On Thu, Dec 27, 2012 at 6:43 PM, ~*~VICKY~*~ wrote:
> I'm giving you a simple recursive code which i wrote long back. Please let
> me know if it fails for any cases. Ignore the funny cout's It used to help
> me debug and i'm lazy to remove it. :P :)
>
> #include
> #include
> using namespace std;
> /*
> abasjc a*c
> while(pattern[j] == '*' text[i] == pattern[j]) {i++; j++}
> */
> bool match(string text, string pattern, int x, int y)
> {
> if(pattern.length() == y)
> {
> cout<<"hey\n";
> return 1;
> }
> if(text.length() == x)
> {
> cout<<"shit\n";
> return 0;
> }
> if(pattern[y] == '.' || text[x] == pattern[y])
> {
> cout<<"in match"< return match(text,pattern,x+1,y+1);
> }
> if(pattern[y] == '*')
> return match(text,pattern,x+1,y) || match(text,pattern,x+1,y+1) ||
> match(text,pattern,x,y+1);
>
> if(text[x] != pattern[y])
> {
> cout<<"shit1\n";
> return 0;
> }
>
> }
>
> int main()
> {
> string text,pattern;
> cin >> text >> pattern;
> cout << match(text, pattern,0, 0);
> }
>
> On Thu, Dec 27, 2012 at 6:10 PM, shady wrote:
>
>> Thanks for the link Ritesh,
>> if (isMatch(s, p+2)) return true;
>> isnt this line incorrect in the code, as it can lead to segmentation
>> fault... how can we directly access p+2 element, we know for sure that p is
>> not '\0', but p+1 element can be '\0' , therefore leading to p+2 to be
>> undefined.
>>
>> On Thu, Dec 27, 2012 at 6:23 AM, Ritesh Mishra wrote:
>>
>>> try to solve it by recursion ..
>>> http://www.leetcode.com/2011/09/regular-expression-matching.html
>>>
>>>
>>> Regards,
>>>
>>> Ritesh Kumar Mishra
>>> Information Technology
>>> Third Year Undergraduate
>>> MNNIT Allahabad
>>>
>>>
>>> On Sun, Dec 23, 2012 at 11:14 PM, Prem Krishna Chettri <
>>> hprem...@gmail.com> wrote:
>>>
Well I can tell you Something about design pattern to solve this case..
What I mean is by using The State Machine Design Pattern, Anyone
can solve this. but Ofcourse it is complicated.
On Sun, Dec 23, 2012 at 11:01 PM, shady wrote:
> that's the point, Have to implement it from scratch... otherwise java
> has regex and matcher, pattern to solve it...
>
>
> On Sun, Dec 23, 2012 at 10:28 PM, saurabh singh
> wrote:
>
>> If you need to implement this for some project then python and java
>> have a very nice library
>>
>>
>> Saurabh Singh
>> B.Tech (Computer Science)
>> MNNIT
>> blog:geekinessthecoolway.blogspot.com
>>
>>
>> On Sun, Dec 23, 2012 at 7:48 PM, shady wrote:
>>
>>>
>>> http://stackoverflow.com/questions/13144590/to-check-if-two-strings-match-with-alphabets-digits-and-special-characters
>>>
>>> any solution for this. we need to implement such regex
>>> tester
>>>
>>> some complex cases :
>>> *string** regex * -> * status*
>>> *
>>> *
>>> reesd re*.d -> match
>>> re*eed reeed -> match
>>>
>>> can some one help with this ?
>>>
>>> --
>>>
>>>
>>>
>>
>> --
>>
>>
>>
>
> --
>
>
>
--
>>>
>>> --
>>>
>>>
>>>
>>
>> --
>>
>>
>>
>
>
>
> --
> Cheers,
>
> Vicky
>
> --
>
>
>
--
Re: [algogeeks] Regex tester
I'm giving you a simple recursive code which i wrote long back. Please let
me know if it fails for any cases. Ignore the funny cout's It used to help
me debug and i'm lazy to remove it. :P :)
#include
#include
using namespace std;
/*
abasjc a*c
while(pattern[j] == '*' text[i] == pattern[j]) {i++; j++}
*/
bool match(string text, string pattern, int x, int y)
{
if(pattern.length() == y)
{
cout<<"hey\n";
return 1;
}
if(text.length() == x)
{
cout<<"shit\n";
return 0;
}
if(pattern[y] == '.' || text[x] == pattern[y])
{
cout<<"in match"<> text >> pattern;
cout << match(text, pattern,0, 0);
}
On Thu, Dec 27, 2012 at 6:10 PM, shady wrote:
> Thanks for the link Ritesh,
> if (isMatch(s, p+2)) return true;
> isnt this line incorrect in the code, as it can lead to segmentation
> fault... how can we directly access p+2 element, we know for sure that p is
> not '\0', but p+1 element can be '\0' , therefore leading to p+2 to be
> undefined.
>
> On Thu, Dec 27, 2012 at 6:23 AM, Ritesh Mishra wrote:
>
>> try to solve it by recursion ..
>> http://www.leetcode.com/2011/09/regular-expression-matching.html
>>
>>
>> Regards,
>>
>> Ritesh Kumar Mishra
>> Information Technology
>> Third Year Undergraduate
>> MNNIT Allahabad
>>
>>
>> On Sun, Dec 23, 2012 at 11:14 PM, Prem Krishna Chettri <
>> hprem...@gmail.com> wrote:
>>
>>> Well I can tell you Something about design pattern to solve this case..
>>>
>>>What I mean is by using The State Machine Design Pattern, Anyone
>>> can solve this. but Ofcourse it is complicated.
>>>
>>>
>>>
>>>
>>> On Sun, Dec 23, 2012 at 11:01 PM, shady wrote:
>>>
that's the point, Have to implement it from scratch... otherwise java
has regex and matcher, pattern to solve it...
On Sun, Dec 23, 2012 at 10:28 PM, saurabh singh wrote:
> If you need to implement this for some project then python and java
> have a very nice library
>
>
> Saurabh Singh
> B.Tech (Computer Science)
> MNNIT
> blog:geekinessthecoolway.blogspot.com
>
>
> On Sun, Dec 23, 2012 at 7:48 PM, shady wrote:
>
>>
>> http://stackoverflow.com/questions/13144590/to-check-if-two-strings-match-with-alphabets-digits-and-special-characters
>>
>> any solution for this. we need to implement such regex
>> tester
>>
>> some complex cases :
>> *string** regex * -> * status*
>> *
>> *
>> reesd re*.d -> match
>> re*eed reeed -> match
>>
>> can some one help with this ?
>>
>> --
>>
>>
>>
>
> --
>
>
>
--
>>>
>>> --
>>>
>>>
>>>
>>
>> --
>>
>>
>>
>
> --
>
>
>
--
Cheers,
Vicky
--
Re: [algogeeks] Regex tester
Thanks for the link Ritesh, if (isMatch(s, p+2)) return true; isnt this line incorrect in the code, as it can lead to segmentation fault... how can we directly access p+2 element, we know for sure that p is not '\0', but p+1 element can be '\0' , therefore leading to p+2 to be undefined. On Thu, Dec 27, 2012 at 6:23 AM, Ritesh Mishra wrote: > try to solve it by recursion .. > http://www.leetcode.com/2011/09/regular-expression-matching.html > > > Regards, > > Ritesh Kumar Mishra > Information Technology > Third Year Undergraduate > MNNIT Allahabad > > > On Sun, Dec 23, 2012 at 11:14 PM, Prem Krishna Chettri > wrote: > >> Well I can tell you Something about design pattern to solve this case.. >> >>What I mean is by using The State Machine Design Pattern, Anyone >> can solve this. but Ofcourse it is complicated. >> >> >> >> >> On Sun, Dec 23, 2012 at 11:01 PM, shady wrote: >> >>> that's the point, Have to implement it from scratch... otherwise java >>> has regex and matcher, pattern to solve it... >>> >>> >>> On Sun, Dec 23, 2012 at 10:28 PM, saurabh singh wrote: >>> If you need to implement this for some project then python and java have a very nice library Saurabh Singh B.Tech (Computer Science) MNNIT blog:geekinessthecoolway.blogspot.com On Sun, Dec 23, 2012 at 7:48 PM, shady wrote: > > http://stackoverflow.com/questions/13144590/to-check-if-two-strings-match-with-alphabets-digits-and-special-characters > > any solution for this. we need to implement such regex > tester > > some complex cases : > *string** regex * -> * status* > * > * > reesd re*.d -> match > re*eed reeed -> match > > can some one help with this ? > > -- > > > -- >>> >>> -- >>> >>> >>> >> >> -- >> >> >> > > -- > > > --
Re: [algogeeks] Regex tester
try to solve it by recursion .. http://www.leetcode.com/2011/09/regular-expression-matching.html Regards, Ritesh Kumar Mishra Information Technology Third Year Undergraduate MNNIT Allahabad On Sun, Dec 23, 2012 at 11:14 PM, Prem Krishna Chettri wrote: > Well I can tell you Something about design pattern to solve this case.. > >What I mean is by using The State Machine Design Pattern, Anyone > can solve this. but Ofcourse it is complicated. > > > > > On Sun, Dec 23, 2012 at 11:01 PM, shady wrote: > >> that's the point, Have to implement it from scratch... otherwise java has >> regex and matcher, pattern to solve it... >> >> >> On Sun, Dec 23, 2012 at 10:28 PM, saurabh singh wrote: >> >>> If you need to implement this for some project then python and java have >>> a very nice library >>> >>> >>> Saurabh Singh >>> B.Tech (Computer Science) >>> MNNIT >>> blog:geekinessthecoolway.blogspot.com >>> >>> >>> On Sun, Dec 23, 2012 at 7:48 PM, shady wrote: >>> http://stackoverflow.com/questions/13144590/to-check-if-two-strings-match-with-alphabets-digits-and-special-characters any solution for this. we need to implement such regex tester some complex cases : *string** regex * -> * status* * * reesd re*.d -> match re*eed reeed -> match can some one help with this ? -- >>> >>> -- >>> >>> >>> >> >> -- >> >> >> > > -- > > > --
Re: [algogeeks] Regex tester
i was asked the same question in my Microsoft interview. I gave the solution using a naive method by comparing each letter and tracking the next character. If the next character is *, then set a flag and check the equality, and if it is a . ignore and move on. He was ok with the solution but said that, there is a better way. On Sun, Dec 23, 2012 at 11:14 PM, Prem Krishna Chettri wrote: > Well I can tell you Something about design pattern to solve this case.. > >What I mean is by using The State Machine Design Pattern, Anyone can > solve this. but Ofcourse it is complicated. > > > > > On Sun, Dec 23, 2012 at 11:01 PM, shady wrote: >> >> that's the point, Have to implement it from scratch... otherwise java has >> regex and matcher, pattern to solve it... >> >> >> On Sun, Dec 23, 2012 at 10:28 PM, saurabh singh >> wrote: >>> >>> If you need to implement this for some project then python and java have >>> a very nice library >>> >>> >>> Saurabh Singh >>> B.Tech (Computer Science) >>> MNNIT >>> blog:geekinessthecoolway.blogspot.com >>> >>> >>> On Sun, Dec 23, 2012 at 7:48 PM, shady wrote: http://stackoverflow.com/questions/13144590/to-check-if-two-strings-match-with-alphabets-digits-and-special-characters any solution for this. we need to implement such regex tester some complex cases : string regex->status reesd re*.d -> match re*eed reeed -> match can some one help with this ? -- >>> >>> >>> -- >>> >>> >> >> >> -- >> >> > > > -- > > --
Re: [algogeeks] Regex tester
@shady : look for the implementation of Matcher class.. may be that could help . On Sun, Dec 23, 2012 at 11:01 PM, shady wrote: > that's the point, Have to implement it from scratch... otherwise java has > regex and matcher, pattern to solve it... > > > On Sun, Dec 23, 2012 at 10:28 PM, saurabh singh wrote: > >> If you need to implement this for some project then python and java have >> a very nice library >> >> >> Saurabh Singh >> B.Tech (Computer Science) >> MNNIT >> blog:geekinessthecoolway.blogspot.com >> >> >> On Sun, Dec 23, 2012 at 7:48 PM, shady wrote: >> >>> >>> http://stackoverflow.com/questions/13144590/to-check-if-two-strings-match-with-alphabets-digits-and-special-characters >>> >>> any solution for this. we need to implement such regex >>> tester >>> >>> some complex cases : >>> *string** regex * -> * status* >>> * >>> * >>> reesd re*.d -> match >>> re*eed reeed -> match >>> >>> can some one help with this ? >>> >>> -- >>> >>> >>> >> >> -- >> >> >> > > -- > > > -- best wishes!! Vaibhav --
Re: [algogeeks] Regex tester
Well I can tell you Something about design pattern to solve this case.. What I mean is by using The State Machine Design Pattern, Anyone can solve this. but Ofcourse it is complicated. On Sun, Dec 23, 2012 at 11:01 PM, shady wrote: > that's the point, Have to implement it from scratch... otherwise java has > regex and matcher, pattern to solve it... > > > On Sun, Dec 23, 2012 at 10:28 PM, saurabh singh wrote: > >> If you need to implement this for some project then python and java have >> a very nice library >> >> >> Saurabh Singh >> B.Tech (Computer Science) >> MNNIT >> blog:geekinessthecoolway.blogspot.com >> >> >> On Sun, Dec 23, 2012 at 7:48 PM, shady wrote: >> >>> >>> http://stackoverflow.com/questions/13144590/to-check-if-two-strings-match-with-alphabets-digits-and-special-characters >>> >>> any solution for this. we need to implement such regex >>> tester >>> >>> some complex cases : >>> *string** regex * -> * status* >>> * >>> * >>> reesd re*.d -> match >>> re*eed reeed -> match >>> >>> can some one help with this ? >>> >>> -- >>> >>> >>> >> >> -- >> >> >> > > -- > > > --
Re: [algogeeks] Regex tester
that's the point, Have to implement it from scratch... otherwise java has regex and matcher, pattern to solve it... On Sun, Dec 23, 2012 at 10:28 PM, saurabh singh wrote: > If you need to implement this for some project then python and java have a > very nice library > > > Saurabh Singh > B.Tech (Computer Science) > MNNIT > blog:geekinessthecoolway.blogspot.com > > > On Sun, Dec 23, 2012 at 7:48 PM, shady wrote: > >> >> http://stackoverflow.com/questions/13144590/to-check-if-two-strings-match-with-alphabets-digits-and-special-characters >> >> any solution for this. we need to implement such regex >> tester >> >> some complex cases : >> *string** regex * -> * status* >> * >> * >> reesd re*.d -> match >> re*eed reeed -> match >> >> can some one help with this ? >> >> -- >> >> >> > > -- > > > --
Re: [algogeeks] Regex tester
If you need to implement this for some project then python and java have a very nice library Saurabh Singh B.Tech (Computer Science) MNNIT blog:geekinessthecoolway.blogspot.com On Sun, Dec 23, 2012 at 7:48 PM, shady wrote: > > http://stackoverflow.com/questions/13144590/to-check-if-two-strings-match-with-alphabets-digits-and-special-characters > > any solution for this. we need to implement such regex > tester > > some complex cases : > *string** regex * -> * status* > * > * > reesd re*.d -> match > re*eed reeed -> match > > can some one help with this ? > > -- > > > --
