Bryan R Harris wrote at Thu, 30 May 2002 01:58:19 +0200:
> I'm trying to come up with an algorithm that seems like it ought to be really easy,
>but it's
> turning out to be pretty tough...
>
> Basically I have three pairs of start/stop times, e.g.:
>
> 3, 5
> 4, 10
> 15, 20
>
> I want the total time covered by all these ranges. I can't just say (5-3 + 10-4 +
>20-15) because
> the 3,5 and 4,10 ranges overlap. The desired answer for this case is 13, 3 to 10
>and 15 to 20.
Sorry, but 3 to 10 are 3,4,5,6,7,8,9,10 = 8 times
and 15 to 20 are 15,16,17,18,19,20 = 6 times.
So the answer should be 8+6 = 14, shouldn't ?
> I
> have the start times in two arrays, @starts and @stops.
>
> I have to do this probably a million+ times. Any ideas on how I could go about this?
>
Of course, the solutions of Beau and Felix are good, TMTWTDI,
heres are very readable solution
(but perhaps very memory reluctant :-( )
my %timings;
$timings{$_}++ for (3 .. 5, 4 .. 10, 15 .. 20);
my $total_time = scalar values %timings;
Greetings,
Janek
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