Yes, but you could solve it when S=2 with 2, 6, and you could not solve it
with 3, 7.

Are you asking if you could solve it with 2, 6, 7 when S=3?  Then yes, of
course, adding tile 7 does nothing to change the fact that 2, 6 was already
a perfectly good solution.

On Wed, 13 Apr 2016 at 14:17 Mohit Soni <[email protected]> wrote:

> On Tuesday, April 12, 2016 at 7:27:38 PM UTC+5:30, Xiongqi ZHANG wrote:
> > > I read it but I think I am missing something to get right...
> > > I just want to know that in Problem D: Fractiles (
> https://code.google.com/codejam/contest/6254486/dashboard#s=p3) at In
> sample case #5, how valid solution are tiles #2 and #6.I mean it could be
> #3 #7 also if we go for maximum occurrence...
> > > It would be a great help if some can make it clear...
> > > Thanks..
> >
> > Checking #2 can verify pos#1 and pos#2
> > Checking #6 can verify pos#2 and pos#3
> >
> > So it is a valid solution
> >
> > Checking #3 can verify pos#1 and pos#3
> > Checking #7 can verify pos#3 and pos#1
> >
> > so pos#2 is not checked, so it is not valid solution.
>
> Is it compulsory ti check pos#2...as S = 3, we can clean 3 tile at a time
> right ?
>
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