4
\$\begingroup\$

You have a phone plan that lets you talk 180 minutes a day. Every day that you go over your daily limit you have to pay a fine of F = Men, where M is the number of minutes you went over and n is the number of days you have gone over (e.g if I went 20 minutes over the daily limit and it's my 2nd day going over, I have to pay 20*e2 = 147.78).

You will be given a list of 12-hour clock times. Even-indexed times represent the start of a phone call, odd-indexed times represent the end of a phone call. Using this list find out how much money the person has to pay in fines.

Notes

  • Please use an approximation of e that is at least as accurate as 2.718
  • You never need to round your answers or include a currency symbol.
  • The first time in the list will always be an A.M (morning) time.
  • Remember that in 12-hour time 1 comes after 12 (you can think of 12 as 0).
  • A day starts at 12:00 AM and ends at 11:59 PM
  • If a number is lower than the previous time, then that means that it is in the opposite (A.M./P.M) time range.
  • A call will never start in one day and end in another.

Test Cases

10:30
2:30
5:30
6:30

This person made two phone calls in one day. One from 10:30 A.M. to 2:30 P.M., another from 5:30 P.M to 6:30 P.M. In total, they talked for 5 hours this day, going 120 minutes over the daily limit. Since it's their first offense, they have to pay 120*e = $326.19 (yes, their phone plan is quite Draconic).


10:30
11:45
5:15
6:45
8:20
9:20
5:30
7:50
2:30
3:30

This person made several phone calls over 2 days. The first day they were on the phone from 10:30-11:45 A.M, 5:15-6:45 P.M., and 8:20-9:20 PM using a total of 225 minutes. That is an overage of 45 minutes. So their fine for the day is 45*e = $122.32.

The next day they talked from 5:30-7:50 A.M and 2:30-3:30 P.M., a total of 200 minutes. They went over by 20 minutes, and since it's their second offense they pay 20*e2 = $147.78

You output their total fee as $122.32 + $147.78 = $270.10


10:30
11:30
6:30
7:30
9:45
11:00
10:25
11:25
8:45
9:45
6:00
8:00
2:30
4:30
  • Day 1:
    • 10:30-11:30 A.M., 6:30-7:30 P.M., 9:45-11:00 P.M. = 195 minutes
    • Fee: 15*e = $40.77
  • Day 2:
    • 10:25-11:25 A.M., 8:45-9:45 P.M. = 120 minutes
    • Fee: $0
  • Day 3:
    • 6:00 - 8:00 A.M., 2:30-4:30 P.M. = 240 minutes
    • Fee: 60*e2 = $443.34

Output: $40.77 + $0 + $443.34 = $484.11

\$\endgroup\$
  • \$\begingroup\$ This sounds a lot like VerizonMath \$\endgroup\$ – Digital Trauma Jan 16 '16 at 22:07
  • \$\begingroup\$ @user8 Yes, as you can see in the first test case. \$\endgroup\$ – geokavel Jan 17 '16 at 1:45
  • \$\begingroup\$ Would the call log 8:00 9:00 7:00 6:00 be parsed as Day 1 8:00AM - 9:00PM, Day 2 7:00AM - 6:00PM or be invalid? If it's valid it would break the rule that "If a number is lower than the previous time, then that means that it is in the opposite (A.M./P.M) time range". \$\endgroup\$ – user81655 Jan 17 '16 at 1:55
  • \$\begingroup\$ @user8 no, the first part is parsed as 8-9 AM, the 2nd part is invalid bcuz it crosses from day 1 to day 2. \$\endgroup\$ – geokavel Jan 17 '16 at 2:00
  • \$\begingroup\$ Oh sorry, I meant to write Day 1 8:00AM - 9:00AM. But thanks for clarifying. \$\endgroup\$ – user81655 Jan 17 '16 at 2:03
1
\$\begingroup\$

Perl 5, 185 bytes

@c=map{s/(.+):(.+)/$1%12*60+$2/er}<>;while(@c){$s=shift@c;$e=shift@c;$p+=0>($k=$e-$s)?720+$k:$k;if($e>$c[0]||$k<0){if($m||!@c){$f+=0>($p-=180)?0:$p*exp++$d;$p=0}$m=!$m}}printf'$%.2f',$f

Try it online!

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.