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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";
+  │
+  └────