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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>
>
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