Re: [algogeeks] interview quest..
if we have RBGGGBGGBR
what should be the answer???
RBGR or (empty string)
On Mon, Feb 7, 2011 at 3:51 PM, rajan goswami rajan.goswam...@gmail.comwrote:
yeh.
Agree with ramkumar.
Simplest solution is to use Stack...
On Sun, Feb 6, 2011 at 8:11 PM, Abhijit K Rao abhijitkrao...@gmail.comwrote:
I could not get it for recursively, but iteratively, I coded a solution.
If anyone knows recursively,
let us know please.
#includestdio.h
void main()
{
char s[18]=DGGDBCBHH;
int i=0,j=0;
int count;
while(s[i]!='\0')
{
if(s[i] == s[i+1])
{
count = strlen(s)-2;
while(count--)
{
s[i]=s[i + 2];
i++;
}
s[i]='\0';
i=0;
}
else
{
i++;
}
}
printf(%s,s);
getch();
}
I/P: DGGDBCBHH O/P: BCB
Best Regards
Abhijit
On Sun, Feb 6, 2011 at 1:47 PM, ramkumar santhanam
ramkumars@gmail.com wrote:
use stack.
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Re: [algogeeks] interview quest..
yeh.
Agree with ramkumar.
Simplest solution is to use Stack...
On Sun, Feb 6, 2011 at 8:11 PM, Abhijit K Rao abhijitkrao...@gmail.comwrote:
I could not get it for recursively, but iteratively, I coded a solution. If
anyone knows recursively,
let us know please.
#includestdio.h
void main()
{
char s[18]=DGGDBCBHH;
int i=0,j=0;
int count;
while(s[i]!='\0')
{
if(s[i] == s[i+1])
{
count = strlen(s)-2;
while(count--)
{
s[i]=s[i + 2];
i++;
}
s[i]='\0';
i=0;
}
else
{
i++;
}
}
printf(%s,s);
getch();
}
I/P: DGGDBCBHH O/P: BCB
Best Regards
Abhijit
On Sun, Feb 6, 2011 at 1:47 PM, ramkumar santhanam
ramkumars@gmail.com wrote:
use stack.
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Re: [algogeeks] interview quest..
use stack. -- 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] interview quest..
I could not get it for recursively, but iteratively, I coded a solution. If
anyone knows recursively,
let us know please.
#includestdio.h
void main()
{
char s[18]=DGGDBCBHH;
int i=0,j=0;
int count;
while(s[i]!='\0')
{
if(s[i] == s[i+1])
{
count = strlen(s)-2;
while(count--)
{
s[i]=s[i + 2];
i++;
}
s[i]='\0';
i=0;
}
else
{
i++;
}
}
printf(%s,s);
getch();
}
I/P: DGGDBCBHH O/P: BCB
Best Regards
Abhijit
On Sun, Feb 6, 2011 at 1:47 PM, ramkumar santhanam
ramkumars@gmail.comwrote:
use stack.
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[algogeeks] interview quest..
can anyone help..? Suppose we have a string RGBBGBGR. we have to eliminate the couple (two same chars adjacent to each other) recursively. For example RGBBGBGR -- RGGBGR--RBGR -- balaji ;-) -- 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.
