@Snehal

just a hint: there is no need of that channel 1 & channel 2.

just treat each program as independent program.

On Mon, Jan 31, 2011 at 10:44 PM, snehal jain <[email protected]> wrote:

> or provide some link
>
>
> On Mon, Jan 31, 2011 at 10:44 PM, snehal jain <[email protected]>wrote:
>
>> @ juver++
>>
>> can you please share your approach
>>
>>
>> On Mon, Jan 31, 2011 at 8:43 PM, Divya Jain <[email protected]>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 <[email protected]> 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 <[email protected]>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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