Count the number of subdirectories without the execute bit

How can I count the number of subdirectories that don’t have the execute bit set?

This is my attempt but is there a better or more elegant way?

count=0; for d in */; do [[ -d $d && ! -x $d ]] && (( ++count )); done
printf %s\\n "$count"

My main interest is in checking if the execute bit is not set for all. That is not just for the current user.

>Solution :

The following counts directories without executable bit for current user:

find . -mindepth 1 -maxdepth 1 -type d '!' -executable -printf . | wc -c

The following counts directories that have 0 executable bits:

find . -mindepth 1 -maxdepth 1 -type d '!' -perm /111 -printf . | wc -c

The following counts directories that have 2 or less executable bits set:

find . -mindepth 1 -maxdepth 1 -type d '!' -perm -111 -printf . | wc -c

Parts from man find:

   -perm /mode                                                                                                                         
          Any  of  the permission bits mode are set for the file.

  -perm -mode
          All  of the permission bits mode are set for the file.

   -executable                                                                                                                         
          Matches files which are executable and directories which are
          searchable (in a file name resolution sense) by the  current
          user.