Files that have at most two ‘a’ characters in their full path - regex

Question

I'm trying to figure out how to find files that have at most two 'a' characters in their full path with AWK.
The following is what I've come up so far but it's not doing the job.
BEGIN{}
{
if( match( $1, ".*[a].*[a].*[^a]+" ) )
print $1
}
END{}
It reads the files names with their full paths from a file called "data" created separately via the following command.
find / -name '*'
What should I modify?

Answer 1

The following is judged too short to be an answer on its own, but it's all I meant to write:
^[^a]*(a[^a]*(a[^a]*)?)?$
By the way, you don't need awk. grep -E would work fine.
But now that I think of it, if you are going to use awk, the following is even simpler:
awk '!/a.*a.*a/'

Answer 2

You have three errors.
You need to include the start-of-line and end-of-line patterns ^ and $ otherwise an arbitrary prefix or suffix may contain some as.
You need to make the occurrences of a optional, by using parenthesis and ?.
.* can contain a so you need to use [^a] to match the non-a characters.
The result would be a regular expression like:
^([^a]*a)?([^a]*a)?[^a]*$
Edit:
As Ed points out in the comments below his answer, if you pass the --re-interval flag to Awk, you can use intervals.
The expression would then be:
^([^a]*a){0,2}[^a]*$
This allows us say we want to find between 0 and 2 as.

Answer 3

The correct solution is this:
awk '!/(.*a){3}/' file
or either of these if your awk doesn't support RE intervals:
awk 'gsub(/a/,"&") < 3' file
awk 'split($0,x,/a/) < 3' file
so in either case if you want to test for fewer than 17 "a"s you just change 3 to 17 (for example):
awk '!/(.*a){17}/' file
rather than writing:
awk '^[^a]*(a[^a]*(a[^a]*(a[^a]*(a[^a]*(a[^a]*(a[^a]*(a[^a]*(a[^a]*(a[^a]*(a[^a]*(a[^a]*(a[^a]*(a[^a]*(a[^a]*(a[^a]*(a[^a]*)?)?)?)?)?)?)?)?)?)?)?)?)?)?)?)?$'
or similar.