nagendra..
plz show the working using an example ..
take input

babbaabbc


On Sun, Sep 6, 2009 at 2:07 PM, Nagendra Kumar <[email protected]>wrote:

>
> @all:
>
>   T(i,j) : denotes the length of longest palindrome with start index
> i  and end index  j index.
>      T(i,j )=
>              max {
>                                 1   ,  if   i == j;
>                                  2+T(i+1,j-1)  if x[i] == x[j];
>                                  max(T(i+1,j) T(i,j-1))
>
>                      }
>   This is the recurrence relation
>    Now algorithm:
>              T(i,i-1) = 0 for 1<= i <= n.
>
>    for i=  1   to n
>       T(i)(i)  = 1;
>     start with the distnace d
>    for d =  1  to n
>     for i = 1 to n -d
>            {    j  = i+d;
>                 if(x[i] == x[j])
>                       T[i][j] = 2+ T[i+1][j-1];
>                else
>                    T[i][j]  = max (T[i+1][j], T[i][j-1])
>
>            }
>
>  return T[1][n];
>
> On Tue, Sep 1, 2009 at 12:01 AM, Dufus<[email protected]> wrote:
> >
> > This is one of the most difficult dynamic programming problem I have
> > faced so far.
> >
> > A good discussion on this problem and different solutions can be found
> > at
> >
> http://www.akalin.cx/2007/11/28/finding-the-longest-palindromic-substring-in-linear-time/
> >
> > _dufus
> >
> >
> > On Aug 31, 6:01 pm, ankur aggarwal <[email protected]> wrote:
> >> @abhijeet
> >> dryrun on this
> >>
> >> abbaabba
> >>
> >> On Mon, Aug 31, 2009 at 5:45 PM, Abhijeet Kumar
> >> <[email protected]>wrote:
> >>
> >> > u can push the string onto a stack ... so last element will be on top
> ...
> >> > den u can pop out element and compare ...
> >> > if u have 2 or more set of palindromes ..
> >> > u can keep a counter and keep a check on dat...
> >
> > >
> >
>
> >
>

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