@atul
The example that u have taken, is it correct ?
I see that in the search string 'abcdtrwdcba' acc to u the even length
palindrome is abcddcba..
On Dec 29, 9:23 am, atul anand <atul.87fri...@gmail.com> wrote:
> @Lucifier :
>
> this is wat i was trying to say :-
>
> string = abcdtrwdcba
>
> find even length substring and hash them , moving from left to right.
>
> hash(abcdtrwdcba) // corner case
> hash(ab)
> hash(abcd)
> hash(abcdtr)
> .
> .
> .
> hash(dcba).
>
> after hashing is done.
>
> again hash moving from right to left.
> hash(abcd) ---> hash alreday present so ...even length palindrome exists.
> you need to take care of cases like for ab ->hash is present , but is part
> of bigger substring (abcd) when moving from right to left. so keep track of
> the index.
> here how to keep track:-
>
> when you find hash of (ab) index=i;
> when you find second even palindrome hash(abcd).
>
> if length("abcd") > length("ab");
> {
> temp_index=current_index + lenght("ab");
> if(temp_index== index)
> {
> // we know ab is the part of bigger string abcd.update
> the new even palindrome found.
> }
>
>
>
>
>
>
>
> }
--
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.
