or provide some link
On Mon, Jan 31, 2011 at 10:44 PM, snehal jain <learner....@gmail.com> wrote:
> @ juver++
>
> can you please share your approach
>
>
> On Mon, Jan 31, 2011 at 8:43 PM, Divya Jain <sweetdivya....@gmail.com>wrote:
>
>> @ above
>>
>> ur code fails for following example
>> channel 1 : prog1 8:00-9:00, prog 2 9:00-10:00
>> channel 2 : prog1 8:15-10:00
>>
>> your code returns 8:15- 10
>> and the answer should be channel1/prog1 + channel1/prog2
>>
>>
>>
>>
>> On 21 January 2011 12:54, Anand <anandut2...@gmail.com> wrote:
>>
>>>
>>> Sort all program with their starting time.
>>>
>>> Appy the below pseudo code to find max number of programs he can watch.
>>>
>>> for(i=i;i<len;i++)
>>> {
>>> /*Check for overlap */
>>> if(p[i].start > p[i-1].end)
>>> {
>>> end = i;
>>> }
>>> else
>>> {
>>> /*Index of the first program to be watch*/
>>> if((p[i-1].end - p[i-1].start) < (p[i].end - p[i].start))
>>> {
>>> start = i;
>>> }
>>> }
>>>
>>> }
>>> return end - start;
>>>
>>>
>>> On Thu, Jan 20, 2011 at 10:11 PM, snehal jain <learner....@gmail.com>wrote:
>>>
>>>> There is a TV avid person. HE wants to spend his max time on TV. There
>>>> are N channels with different program of different length and diff
>>>> times. WAP so that the person cam spend his max time watching TV.
>>>> Precondition: If that person watches a program, he watches it
>>>> completely.
>>>>
>>>> Ex:
>>>> Channel1: prog1 – 8:00- 8:30
>>>> prog2: 9:00 – 10:00
>>>> prog3: 10:15 – 12:00
>>>>
>>>> channel2: prg1 – 8:15 – 10:00
>>>> prg2: 10:30 – 12:00
>>>>
>>>> So in this case max time will be if he watches:
>>>>
>>>> ch2/prg1 + ch1/prg3
>>>>
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