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━━━━━━━━━━━━━━━
CHALLENGE 086
Andinus
━━━━━━━━━━━━━━━
2020-11-15
Table of Contents
─────────────────
1. Task 1 - Pair Difference
.. 1. Perl
1 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.
1.1 Perl
────────
• Program: <file:perl/ch-1.pl>
@N & $A are taken from stdin, $A is the last argument.
┌────
│ die "usage: ./ch-1.pl <integers \@N> <integer \$A>\n"
│ unless scalar @ARGV >= 3;
│
│ my $A = pop @ARGV;
│ my @N = @ARGV;
└────
We just loop over @N over a loop over @N & find the difference. If
it's equal to `$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.
┌────
│ 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";
│
└────
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