1 pass clean solution.
Intuition\nIf two letters are really close in the alphabet (like \'a\' and \'b\' or \'x\' and \'y\'),increment count. Then it moves to the next pair of letters 
# Intuition\nIf two letters are really close in the alphabet (like \'a\' and \'b\' or \'x\' and \'y\'),increment count. Then it moves to the next pair of letters and does the same thing until it finishes checking all pairs.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n- O(N), N is the input length\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n- O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public int removeAlmostEqualCharacters(String word) {\n        int cnt = 0;\n\n        for (int i = 1; i < word.length();) {\n            char prev = word.charAt(i-1);\n            char curr = word.charAt(i);\n            if (Math.abs(curr - prev) <= 1) {\n                cnt++;\n                ++i;\n            }\n            ++i;\n        }\n        return cnt;\n    }\n}\n```
['Java']