Anagram function returns False for unknown reason in some words - Python

I've tried to create a function that returns True if two words are anagrams (have the same letters) to each other.

I realized that my counter counts to 13 but not sure what is the issue.

def is_anagram (worda: str , wordb: str) -> bool:
    count_same_letters = 0
    if len(worda) == len(wordb):
        for i in worda:
            for j in wordb:
                if i == j:
                    count_same_letters = count_same_letters+1
                    print(count_same_letters)
                    print(len(wordb))
        return count_same_letters == len(wordb)
    else:
        return False

print(is_anagram("anagram","nagaram"))

while trying the string 'abc' abd 'bca' the output was True as I expected, but the strings 'anagram' and 'nagaram'returns False

As mentioned, the problem here is that letters occurring more than once are counted again. Supposing you need to implement the algorithm yourself without Count, here's one approach to the problem:

def is_anagram (worda: str , wordb: str) -> bool:
    if len(worda) == len(wordb):
        remaining_letters = wordb
        for letter in worda:
            letter_index = remaining_letters.find(letter)
            if letter_index == -1:
                return False # no! not anagram
            
            remaining_letters = remaining_letters[:letter_index] + remaining_letters[letter_index+1:]
        
        if len(remaining_letters) == 0:
            return True
    else:
        return False
                
print(is_anagram("anagram", "gramnaa"))
print(is_anagram("omg", "omggg"))
print(is_anagram("abc", "cba"))
print(is_anagram("aaaa", "aaaaaa"))  

By "deleting" every letter we find from remaining_letters our code will have to search through a smaller string on each iteration, possibly making it perform faster then some other alternatives.