统计一个字符串在该字符串的反序position中出现的次数

Question

I am trying to figure out a way that you can count how many times a letter appears in the exact opposite position. For example:

word = 'ABXCEEVHBA' --> The correct output give me 3 because A is first and last. B is second and second from last and so forth.

I have found an answer that gives me the correct result but I was wondering if there is a more elegant way to do this ideally with no modules.

word = 'ABXCEEVHBA'
reverse = ''.join(reversed(word))

sum =0
for i in range(len(word)):
    if word[i]==reverse[i]:
        sum+=1

print(int(sum/2))

Answer 1

Believe this shall do it:

>>> count = 0
>>> for i in range(len(word)//2):  # meet half-way.
    if word[i] == word[~i]:
        count += 1

        
>>> count
3

Answer 2

You can do it with zip by combining the string with the inverse of itself:

sum(a==b for a,b in zip(word,reversed(word)))//2