Let's say you have two lists, @a and @b, and all elements of @b are contained in @a. How do you get "@a without @b"? Of course, one could sort both lists, iterate through the elements of @a, have an index for @b, and push elements in @a to an array @c if the element in @a is not at that position in @b. But that seems rather tedious. For instance, 
@b = sort {$a <=> $b} @b;
@a = sort {$a <=> $b} @a;
print "@b\n";

for (@a) {
  if (defined($b[$i])) {
    if (not $_ == $b[$i]) {
      push @c, $_
    } else {
      $i++
    }
  } else {
    push @c, @a[($j..$#a)];
    last;
  }
  $j++;
}

[download]

I'm sure this can be done in a somewhat more efficient (perl-ish *g) way ... like this: 
@c = grep { 
  defined($b[0]) and !($b[0] eq $_) or !shift @b
} sort {$a <=> $b} @a;

[download]

This should work if @b is already sorted. I'm sure it can be done in better ways .. I'm brand-new at Perl after all.

John

In reply to most efficient way to implement "A without B" by telcontar

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