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 > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.