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#+SETUPFILE: ~/.emacs.d/org-templates/level-2.org
#+HTML_LINK_UP: ../index.html
#+OPTIONS: toc:2
#+EXPORT_FILE_NAME: index
#+DATE: 2020-11-15
#+TITLE: Challenge 086
* Task 1 - Pair Difference
You are given an array of integers @N and an integer $A.
Write a script to find find if there exists a pair of elements in the
array whose difference is $A.
Print 1 if exists otherwise 0.
** Perl
- Program: [[file:perl/ch-1.pl]]
@N & $A are taken from stdin, $A is the last argument.
#+BEGIN_SRC perl
die "usage: ./ch-1.pl <integers \@N> <integer \$A>\n"
unless scalar @ARGV >= 3;
my $A = pop @ARGV;
my @N = @ARGV;
#+END_SRC
We just loop over @N over a loop over @N & find the difference. If it's
equal to =$A= or =-$A= then we print =1= & exit. The first loop is =shift='ing
the numbers out of array =@N= because we are matching for both =$A= & =-$A= so
we don't need the number again.
For example, if =@N = [1, 2]= & we don't =shift= in first loop then we'll
perform 2 subtraction operations: =1 - 2= & =2 - 1= & we won't have to match
for =-$A= but if we just match for =-$A= then we can use =shift= & we'll only
have to perform 1 subtraction operation =1 - 2=.
We assume subtraction costs more than matching with =-$A=, that makes this
more efficient. But it doesn't matter much.
#+BEGIN_SRC perl
while (my $int = shift @N) {
foreach (@N) {
my $diff = $int - $_;
print "1\n" and exit 0 if ($diff == $A or $diff == -$A);
}
}
print "0\n";
#+END_SRC
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