no this is wrong. maintain 2 arrays

int max_calls[no of cust][31]
int current_no_of_calls[no of customers]

both array of customers are initialized to zero.

on call
current_no_of_calls[cust_id]++;
if above > max[id][day] then max_calls[id][day] = above

on hangup
current_no_of_calls[cust_id]--;

On Tue, Mar 29, 2011 at 4:29 PM, Ashim Kapoor <[email protected]> wrote:

>
> int max_calls[no_of_customers][30];
>
> On any phone call --> max_calls[customer_id][day]++;
> On hangup --> max_call[customer_id][day]--;
>
> This would store max calls for each customer on each day. Does the length
> of the call have to be taken into account ? Your question is not clear on
> that.
>
>
> On Tue, Mar 29, 2011 at 3:51 PM, bittu <[email protected]> wrote:
>
>> Pretend you work for a phone company. At your company, yoy u have a
>>
>> satellite that routes phone calls. We want to bill customers by the
>> maximum number of simultaneous phone calls they make in a single day.
>> ( clarifying questions with  the following information: assume no
>> calls last more than 24 hours and that at midnight each night all the
>> calls are automatically dropped. In the event that one call ends as
>> soon as another starts).
>>
>> What information should the satellite store for each phone call?
>> Define a data structure for this (e.g. write a struct).
>> Write a function that finds the maximum number of simultaneous phone
>> calls from a given customer.
>>
>> (Hint: typical solution is O(nlogn), but if you use an absurd amount
>> of memory  it can be done in O(n)).
>>
>> Best Designing & DS, Approach Will b highly Appreciated
>>
>> Thanks
>> Shashank
>>
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