Now that you mention it I agree with you, though I have not realised it earlier. Sorting is an overkill when just min value is required. But hey we cover every angle in our homework service :)

In which case your List::Util::reduce solution or a basic min-find will suffice:

       $idx = $arr[0];
       for(0..$#arr) {
                $idx = $_ if( ($arr[$idx]<0) || (($arr[$_] >= 0) && ($
+arr[$_] < $arr[$idx])) )
        }
[download]

$idx = $arr[0];
for(0..$#arr) {
        $idx = $_ if( ($arr[$idx]<0) || (($arr[$_] >= 0) && ($arr[$_] 
+< $arr[$idx])) )
}
[download]

This returns strange results for certain corner cases:

c:\@Work\Perl\monks>perl -wMstrict -le
"use Data::Dump qw(pp);
 ;;
 for my $ar (
   [ ], [ -1 ], [ -99 ], [ 0, -99 ], [ -99, 0 ],
   ) {
   my $idx = $ar->[0];
   for (0..$#$ar) {
     $idx = $_ if( ($ar->[$idx]<0) || (($ar->[$_] >= 0) && ($ar->[$_] 
+< $ar->[$idx])) );
     }
   print qq{(@$ar) -> lnn index of }, pp $idx;
   }
"
() -> lnn index of undef
(-1) -> lnn index of 0
Use of uninitialized value in numeric lt (<) at -e line 1.
(-99) -> lnn index of -99
(0 -99) -> lnn index of 0
Use of uninitialized value in numeric lt (<) at -e line 1.
Use of uninitialized value in numeric lt (<) at -e line 1.
Use of uninitialized value in numeric lt (<) at -e line 1.
(-99 0) -> lnn index of -99
[download]

I'd hate to have to extend this to dealing with fractional numbers. The O(n) solutions here and here both return an unambiguous undef for lists containing no LNN, including empty lists.

---

Give a man a fish:  <%-{-{-{-<

Sure! I have somehow missed the O(n) solution you provided very early on. Had I seen it, I would have referred to it as it is more complete than my sketch.

I have voted only for your solution. Because while everyone is showing solutions with sort, map grep. I just don't understand that after 30 posts finally someone is talking some sense.

If I would have needed to solve something like this, I would have come up with a simple loop. Job done, move on:

use strict ;
use warnings ;

my @arr = ( 3, 4, 71, 1, 1.5, -598, -100203, 0.5, -2, -1.5 );

my $result ;
while (my ($ix, $val) = each @arr) {
    next if ( $val < 0 || int $val != $val || ($result && $val > $resu
+lt->[0]) ) ;
    $result = [ $val, $ix ] ;
}
if ( $result ) {
    print $result->[0] . " at " . $result->[1] . "\n" ;
}
__END__
1 at 3
[download]

Congratulations: Another O(n) solution! Please note that the each ARRAY syntax is only available from Perl version 5.12 onward.

---

Give a man a fish:  <%-{-{-{-<