Simple Perl Interview Question, Perl

Simple Perl Interview Question

#perl 5.22.1

# find all common numbers in array1 and array2 and print them out
@array1 = (4, -1, 5, 3, 9, -2, 1, 0, 10, 6);
@array2 = (8, 3, -4, 5, 2, 1, 7, -1, -9);

# Method 1 foreach in array2 lookup in array1
# Time complexity T(N*M)
print "Method 1\n";
foreach $num2 (@array2) {
    foreach $num1 (@array1) {
        if ($num2 == $num1) {
            print "$num2\n";
        }
    }
}

# Method 2 sort array1 and array2, scan from begining to the end
# Time complexity O(NlogN + MlogM + N + M)
print "Method 2\n";
@sorted1 = sort {$a <=> $b} @array1;
@sorted2 = sort {$a <=> $b} @array2;
$i = $j = 0;
while ($i <= $#array1 && $j <= $#array2) {
    if ($sorted1[$i] == $sorted2[$j]) {
        print "$sorted1[$i]\n";
        $i++;
        $j++;
    } elsif ($sorted1[$i] < $sorted2[$j]) {
        $i++;
    } else {
        $j++;
    }
}
        
# Method 3 using hashmap
# Time complexity O(N + M)
print "Method 3\n";
%dict = ();
for $num1 (@array1) {
    $dict{$num1} = 1;
}
for $num2 (@array2) {
    if (defined($dict{$num2})) {
        print "$num2\n";
    }
}


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