Comparing all elements with all elements would be quadratic.
Here's a solution that runs in time
O (n log n + k)
where
n is the number of rows and
k
the number of pairs for which the fields 0, 2 and 3 are equal.
This could still be inefficient, k could be
n^2/2 but still nothing is reported because we only
need to report the pairs for which field 1 differs. OTOH, if
it's known that if the fields 0, 2, and 3 are equal that then
field 1 is different, the algorithm used is optimal:
use strict;
use warnings;
my @AoA;
while (<DATA>) {
push @AoA => [split];
}
my @sort = sort {$a -> [0] cmp $b -> [0] ||
$a -> [2] <=> $b -> [2] ||
$a -> [3] <=> $b -> [3]} @AoA;
foreach my $i (0 .. $#sort - 1) {
foreach my $j ($i + 1 .. $#sort) {
last unless $sort [$i] -> [0] eq $sort [$j] -> [0] &&
$sort [$i] -> [2] == $sort [$j] -> [2] &&
$sort [$i] -> [3] == $sort [$j] -> [3];
next if $sort [$i] -> [1] eq $sort [$j] -> [1];
print "[@{$sort[$i]}] and [@{$sort[$j]}]\n";
}
}
__DATA__
AAA BUY 98 0
BBB SEL 27 1
FFF BUY 43 4
AAA SEL 98 0
CCC SEL 98 0
Abigail
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