Python3 -- Easy Solution -- Beats 100% 
IntuitionUsing a Sliding Window of size two and saving a set of the position we updated as the res (so if we need to update it twice it doesn't count more than
# Intuition
Using a **Sliding Window** of size two and saving a set of the position we updated as the res (so if we need to update it twice it doesn't count more than once in the result)
# Code
```python3 []
class Solution:
    def removeAlmostEqualCharacters(self, word: str) -> int:
        res = set() 
        word_list = [ord(w) for w in word.lower()] # so we can use ascii easily
        for index in range(len(word)-1):
            if abs(word_list[index]- word_list[index+1])<=1:
                target = index +1
                if index in res:
                    target -= 1
                word_list[target] += 2
                res.add(target) 
        return len(res)

           
        
        
```
['Python3']