Vaibhav,
Ok write your code and paste on ideone. It should be easy and quick to code
:)


On Fri, Jul 22, 2011 at 7:59 PM, <[email protected]> wrote:

> U hv got my algo completely wrong. Gimme a smaller test case so that i may
> wrk it out fr u. This one is freakingly large :P.
>
>
> Regards
> Vaibhav Mittal
> Computer Science
> Netaji Subhas Institute Of Technology
> Delhi.
>
> On , Pankaj <[email protected]> wrote:
> > abcdeaifjlbmnodpq
> > For this once you encounter a at 6th position, You can update your
> max.Now You will have to do following operation.
> > First clear all the hash.
> > 2. You now can not start from 6th position. You will have to do back and
> start from 2 position that is b.
> >
> >
> >
> >
> > Right?
> > What is the maximum unique string in above test case?
> > Try printing each sub string formed whenever you hit a duplicate.
> > It's still O(N) though if we have constant number of characters :)
> >
> >
> >
> >
> > On Fri, Jul 22, 2011 at 7:50 PM, [email protected]> wrote:
> >
> >
> > Have a look again. I traverse the string just once performing updation on
> variables low, high, max. I assume array operations to be O(1) (which they
> are). OVerall complexity is O(n).Once I hit a duplicate i change my low,
> high and A accordingly and move forward.
> >
> >
> >
> > Regards
> > Vaibhav Mittal
> > Computer Science
> > Netaji Subhas Institute Of Technology
> > Delhi.
> >
> >
> > On , Pankaj [email protected]> wrote:
> >
> >
> > > VaibhavWhat do you think the complexity of your algo is. And once you
> hit an duplicate, from where will you start again?
> > > Try for this example
> > > abcdeaifjlbmnodpq.
> > > It will be still O(26*n) as at max, we would have to start from each
> letter and go forward maximum 26times( if we reach 26 then we have it
> largest possible unique string). Otherwise keep checking.
> >
> >
> > >
> > >
> > > So overall maximum complexity can be (MAX*n) where max will be unique
> letters.
> > > Abhishek, I think the final complexity can never be O(n^2) if you only
> consider unique letters a-z.
> > > The suggested soln is purely bruteforce. If anyone can think of a
> better solution then please reply.
> >
> >
> > >
> > >
> > >
> > >
> > > Cheers
> > > Pankaj
> > >
> > > On Fri, Jul 22, 2011 at 7:37 PM, Pankaj [email protected]>
> wrote:
> >
> >
> > >
> > >
> > >
> >
> > > VaibhavWhat do you thing the complexity of your algo is. And once you
> hit an duplicate, from where will you start again?
> >
> > >
> > >
> > >
> >
> >
> > >
> > >
> > > On Fri, Jul 22, 2011 at 7:21 PM, [email protected]> wrote:
> > >
> > >
> > > String is "abcded"
> > > l =0, h = 0
> >
> >
> > > i = 1, l = 0, h = 1, max = 1, A[a]=1
> > > i = 2, l = 0, h = 2, max = 2, A[b] = 2
> > >
> > >
> > >
> > >
> > > i = 3, l = 0, h = 3, max = 3, A[c] = 3
> > > i = 4, l = 0, h = 4, max = 4, A[d] = 4
> >
> >
> > > i = 5, l = 0, h = 5, max = 5, A[e] = 5
> > > i = 6, 'd' is encountered again, update l = A[d] = 4, new A[d] = 5, h =
> 6, max = max(5, 6-4)= max(5, 2) = 5
> > >
> > >
> > >
> > >
> > >
> > > hence ur ans = 5
> >
> >
> > >
> > > Regards
> > > Vaibhav Mittal
> > > Computer Science
> > > Netaji Subhas Institute Of Technology
> > > Delhi.
> > >
> > >
> > > On , Interstellar Overdrive [email protected]> wrote:
> >
> >
> > >
> > >
> > >
> > >
> > > > @svm11: Take the case with original string "abcded" output should be
> 5 but your algo will give the answer as 0.
> > > >
> > > >
> > > >
> > > >
> >
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