Could you explain the solution for shere problem
On Thu, Aug 25, 2011 at 3:49 PM, vikas <vikas.rastogi2...@gmail.com> wrote:
> 5th qs
>
> r = R(3-2sqrt(2))
>
> On Aug 25, 1:56 pm, vikas <vikas.rastogi2...@gmail.com> wrote:
> > @ All,
> > 1. build a interval tree using startpoints as the key
> > 2. augment this tree such that each interval contains the number of
> > ppl arrived, in this case 1.
> > 3. use this tree and traverse , use this check, if start/end of tree
> > node is inbetween the interval you are searching, person was there.
> >
> > sample code
> > getMaxNumPpl(node *root, int start, int end)
> > {
> > int n = 0;
> > if(root == NULL) return 0;
> > if(CHECK(root->start, root->end, start, end)){
> > n++;
> > }
> > n += getMaxNumPpl(root->left, start, end);
> > n += getMaxNumPpl(root->right, start, end);
> > return n;
> >
> > }
> >
> > On Aug 24, 8:42 pm, rakesh kumar <rockey.rav...@gmail.com> wrote:
> >
> >
> >
> >
> >
> >
> >
> > > Hi
> >
> > > Anybody has answer for sphere problem...could you please proivde
> >
> > > On Wed, Aug 24, 2011 at 9:10 PM, rakesh kumar <rockey.rav...@gmail.com
> >wrote:
> >
> > > > Hi All,
> > > > for checkouts problem how about finding the median for all the
> times....
> >
> > > > 8-00 8-15 830
> > > > sort the second list
> > > > 8-30 900 920
> > > > if we take the mediun of whole list then it will be 8-30 where max
> no of
> > > > people will be present
> >
> > > > Will it work..
> >
> > > > Any body has any idea??
> >
> > > > On Wed, Aug 24, 2011 at 12:58 AM, DK <divyekap...@gmail.com> wrote:
> >
> > > >> Well, strictly speaking, you don't need any complex data structures:
> >
> > > >> *1. Create an array of entities*
> > > >> eg. Person data[100];
> > > >> where
> > > >> struct Person {
> > > >> .... // Person data
> > > >> };
> >
> > > >> *2. Create an array of timestamps:*
> > > >> Event time[200]; // Note: double the size of the Person data array.
> One
> > > >> for start and one for finish time.
> > > >> where
> > > >> struct Event {
> > > >> Person *p; // To maintain a reference to the original person data
> > > >> int time; // say in seconds
> > > >> bool finish; // default: false
> > > >> };
> >
> > > >> *3. Sort the time array based on the time value*
> >
> > > >> *4. Now, simply maintain a counter*
> > > >> int num_people = 0;
> > > >> int max = 0;
> > > >> for each event in time:
> > > >> if(event.finish == true) --num_people;
> > > >> else ++num_people;
> > > >> if(num_people > max) max = num_people;
> >
> > > >> Time Complexity: O(N log N)
> > > >> Space Complexity: O(N)
> >
> > > >> --
> > > >> DK
> >
> > > >>http://twitter.com/divyekapoor
> > > >>http://www.divye.in
> > > >>http://gplus.to/divyekapoor
> >
> > > >> --
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