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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