No, combining bits and pieces of RudiC's suggested code, and CarloM's (or my corrected version of CarloM's) code gives you a hybrid solution that does runs slower and only considers the contents of one file.

I would still suggest my solution (slightly modified to use your new set of files):

Code:

awk -F '[.]' '
$3 > m3 {m3 = $3; m4 = $4; next}
$3 == m3 && $4 > m4 {m4 = $4}
END {print "xxx.xxx." m3 "." m4}
' /Users/Manu/files/*

but I may be biased.

Note that if you did want to sort a list of files in the directory /usrs/Manu/files, you would have to leave off the double quotes in the command:

Code:

sort /Users/Manu/files/*

With the command:

Code:

sort "/Users/Manu/files/*"

it will only sort the single file named * in that directory or produce a file not found diagnostic if that file doesn't exist.