Just another way to do it without loops (kind of cheating):
my $c=0;
s/ \( (?{++$c})
| \) (?{--$c})
| ([^()]+)
/$1 if ($c==0)/gex;
If you match an open paren, increment a counter;
if you match a close paren, decrement it;
if you match a string of non-parens, capture it;
substitute whatever was captured back in if the counter is zero.
There is an "if", but it can be replaced by an "x" operator, if you're a stickler. Or you could rewrite the replacement as $c==0 and $1.
Another little variation that doesn't rely on the (?{}) construct:
s{([()]*)([^()]+)}
{$c += $1=~y/(// - $1=~y/)//;
$2 x !$c
}ge;
Match any parens followed by any non-parens;
Add the count of opens and subtract the count of closes from your counter;
Substitute the non-paren portion back in if the counter is zero.
We're not really tightening our belts, it just feels that way because we're getting fatter.
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