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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