This is just a conversion to Perl of this pseudocode. It finds a Longuest Ascending Sequence in O(N log N). The sequence it happens to find is the first sequence when sorted numerically.

# Minimal Longest Ascending Subsequence
sub mlas {
my $n_las; # Length of output sequence
my @las; # Output sequence
my @terminals; # Array of terminals
my @backptrs; # Array of back pointers
$terminals[1] = 0;
$backptrs[0] = -1;
$n_las = 1;
for my $i (1..$#_) {
my $low = 1;
my $high = $n_las;
while ($low <= $high) {
my $mid = int(($low + $high) / 2);
if ($_[$i] <= $_[$terminals[$mid]]) {
$high = $mid - 1;
} else {
$low = $mid + 1;
}
}
$terminals[$low] = $i;
if ($low <= 1) {
$backptrs[$i] = -1;
} else {
$backptrs[$i] = $terminals[$low - 1];
}
if ($low > $n_las) {
$n_las++;
}
}
for (
my $i = $terminals[$n_las];
$i != -1;
$i = $backptrs[$i]
) {
unshift(@las, $_[$i]);
}
return @las;
}
{
my @a = (8, 6, 5, 1, 9, 3, 7, 4, 2, 10);
my @mlas = mlas(@a);
local $, = ", ";
local $\ = "\n";
print(@mlas); # 1, 3, 4, 10
}
{
my @a = (3, 10, 6, 1, 5, 7, 8, 2, 4, 9);
my @mlas = mlas(@a);
local $, = ", ";
local $\ = "\n";
print(@mlas); # 1, 5, 7, 8, 9
}

The page mentions a secant search would be faster than a binary search although the O() of would still be the sameThe O() would be slightly better, as noted by jdalbec.