String validation with multiple levels required using CMD command - validation

Question

I have three requirements I need to meet when validating a password. I have figure out with the help of others how to verify that the password is at least seven characters long and the user name is not part of the password.
My last requirement is to check to see if a string contains characters from three of the following four groups:
English uppercase characters (A through Z)
English lowercase characters (a through z)
Base 10 digits (0 through 9)
Non-alphabetic characters (for example, !, $, #, %)
For example (Hou$e or House1) would pass but (House, house or hou$e) would fail
The call to ":checkRequirement3" is where I would like to make this finial check. The password is valid if all three requirements are meet.
#echo off
setlocal
set /p userName=Username:
set /p userPassword=Password:
call :strlen result userPassword
call :checkRequirement1
call :checkRequirement2
call :checkRequirement3
ECHO Finished
Pause
:strlen <resultVar> <stringVar>
REM THIS DETERMINES THE LENGTH OF THE PASSWORD
(
setlocal EnableDelayedExpansion
set "s=!%~2!#"
set "len=0"
for %%P in (4096 2048 1024 512 256 128 64 32 16 8 4 2 1) do (
if "!s:~%%P,1!" NEQ "" (
set /a "len+=%%P"
set "s=!s:~%%P!"
)
)
)
(
endlocal
set "%~1=%len%"
exit /b
)
:checkRequirement1
REM THIS CHECKS IF PASSWORD IS AT LEAST 7 CHARACTERS LONG
(
if %result% GEQ 7 (
exit /b
)else (
GOTO passwordFail
)
:checkRequirement2
REM THIS CHECKS IF THE USER NAME IS INCLUDED IN THE PASSWORD
setlocal enabledelayedexpansion
set replacedUsername=!userPassword:%userName%=!
if not !replacedUsername!==%userPassword% (
GOTO passwordFail
)else (
exit /b
)
:checkRequirement3
REM THIS CHECKS IF THE PASSWORD CONTAINS CHARACTERS FROM 3 OF THE FOLLOWING GROUPS
REM English uppercase characters (A through Z)
REM English lowercase characters (a through z)
REM Base 10 digits (0 through 9)
REM Non-alphabetic characters (for example, !, $, #, %)
ECHO This Requirement not finished
Pause
exit /b
:passwordFail
ECHO Password Failed Requirement
PAUSE
exit /b
)

Answer 1

#ECHO OFF
SETLOCAL
FOR %%t IN (Hou$e House1 House, house hou$e hou%%%%se hou^^se) DO CALL :test "%%t"
GOTO :EOF
:test
SET /a count=0
>"q43120516.txt" ECHO %~1
TYPE "q43120516.txt"
FOR %%s IN ("[ABCDEFGHIJKLMNOPQRSTUVWXYZ]"
"[abcdefghijklmnopqrstuvwxyz]"
"[!##&$%%^]"
"[0123456789]") DO FINDSTR /r %%s "q43120516.txt">nul&IF NOT ERRORLEVEL 1 SET /a count+=1
ECHO found %count% groups IN %~1
DEL "q43120516.txt"
GOTO :EOF
naturally, the name of the temporary file "q43120516.txt" is irrelevant.
A few little things to note here:
Certain characters, like % and ^ which have a special meaning to cmd need to be doubled - sometimes quadrupled.
Yes, I'm aware that in theory you could use echo %~1|. Try it.
Yes, I'm aware that in theory you could use [A-Z]. Try it.

Answer 2

This will correctly verify that the password is using 3 out of the 4 groups.
setlocal enabledelayedexpansion
SET /P userPassword=UserPassword:
SET /P userPassword=UserPassword=%userPassword:&=^&%
SET /a count=0
echo %userPassword:&=^&% | findstr /R /C:"[ABCDEFGHIJKLMNOPQRSTUVWXYZ]">null&if not errorlevel 1 SET /a count+=1
echo %userPassword:&=^&% | findstr /R /C:"[abcdefghijklmnopqrstuvwxyz]">null&if not errorlevel 1 SET /a count+=1
echo %userPassword:&=^&% | findstr /R /C:"[!##&$%%^]">null&if not errorlevel 1 SET /a count+=1
echo %userPassword:&=^&% | findstr /R /C:"[0123456789]">null&if not errorlevel 1 SET /a count+=1
If %count% GEQ 3 (
echo Password Meets requirement
exit /b
) else (
goto passwordFail
)