Gentle Punch Card Don,

Sorting is order N-LogN complexity. For very large arrays where you care about efficiency, there should be linear methods where you avoid the sort.

If the numbers in the arrays were multi-digit and there were no repeats in either array, I'd suggest an approach like

• Make each element of @array1 a key for a hash with any corresponding value.
• Foreach element in array2, check to see if it exists.
• If not, report failure.
• If so, delete.
• At the end, the hash should be empty.

You've said in a later note that the numbers were single digit. Then either the arrays are short or there are duplicates. Maybe you don't care about complexity. But maybe you do; I'd modify the algorithm to something like

my %vhash;
foreach (@array1){                 # Each $vhash{N} says how many time
+s N showed up in array1
  $vhash{$_}++};
foreach (@array2){                # Each appearance of N in array2 kno
+cks down its count in vhash
  if(! exists $vhash{$_}){        
    print "Found too many $_ in array2";
    exit}
  elsif($vhash{$_}==1){
    delete $vhash{$_)}
  else{$vhash{$_}--};
if(keys %vhash){
  print "Found too many ",
     join(' ', keys %vhash),
     "in array1\n"}
else{
  print "Yep, they matched!\n"}
[download]

(not tested)

throop

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