Input: n meetings
A meeting e(i) has start time s(i) and finish time f(i).
Sort n events based on there finish time f(i)
Say sorted meeting(based on finishing time) are:
e1,e2,e3,e4,...en

E(t): denotes the maximum no of meetings conducted where 0 <= time <= t.

E(f1) = e1
E(f2) = max(E(s2) + 1), E(f1));
E(f3) = max(E(s3)+1), E(f2));

Make array of E based on finishing times.
So your array will have values of E at different finishing times
f1 E1
f2, E2
f3 E3
f4 E4

so E2 means maximum no of events between time >= f2 and < f3.

Hope it is now clear.

On Wed, Feb 9, 2011 at 12:59 PM, Sachin Agarwal
<sachinagarwa...@gmail.com>wrote:

> didn't quite get it, can you please elaborate.
> thanks
>
> On Feb 8, 10:38 pm, Ujjwal Raj <ujjwal....@gmail.com> wrote:
> > Use dynamic programming:
> > 1) Sort the events in order of finishing time.
> > f1, f2, f3, f4, ... fn
> >
> > E(fn) = E(sn) + 1;
> > Solve the above recursion
> > sn is start time of event n
> >
> > On Wed, Feb 9, 2011 at 11:30 AM, Sachin Agarwal
> > <sachinagarwa...@gmail.com>wrote:
> >
> >
> >
> >
> >
> >
> >
> > > Suppose I have a room and I want to schedule meetings in it. You're
> > > given a list of meeting start and end times. Find a way to schedule
> > > the maximum number of meetings in the room.
> >
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