@Manish Well Explained:)
@sulekha Probably you have to read Head First java which is an awesome
book that covers these topics with examples
On 12 March 2013 21:10, manish untwal wrote:
> according to my understanding the super class object have limited function
> like for example
> Class - ani
hi
> void XYZ(int a[],int b[], int c[])
> {
> int i,j,k;
> i=j=k=0;
> while((i {
> if (a[i] c[k++]=a[i++];
> else
> c[k++]=b[j++];
>0
> }
In this case either i value or j value is incremented at a time
in an iteration . So its impossible that both the conditions (i< n)
and (j
> Which of
m, you will need to have some more checks.
These checks are enough . It will work in all cases
Regards
Sasi kumar T
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@goog
Hi
> User space applications does not have access to physical address. Its just
> the Virtual Address gets printed.
>
+1 on this . The logical to physical memory mapping is handled by
kernel . Only logical address is visible to user
Regards
Sasi kumar T
--
You received thi
hi
> then wat is logical address?
It is just a relative address . Not the exact memory location on RAM
Regards
Sasi kumar T
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@goo
. If a character occurs for an odd number of time
then it will not be a palindrome. Check for even number of
occurences of every character. If the length of the string is odd .
One character must occur odd number of time .
So the problem can be solved just by counting
Regards
Sasi kumar T
-
