BEATS 100 % || HEAVILY COMMENTED
Intuition\n Describe your first thoughts on how to solve this problem. \nThe goal is to determine the minimum number of operations required to remove adjacent a
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal is to determine the minimum number of operations required to remove adjacent almost-equal characters. By iterating through the string and checking for equality or adjacency in the alphabet, we can identify and handle pairs of almost-equal characters. Incrementing the counter for each such pair ensures that we account for the necessary operations. The use of a StringBuilder allows us to make modifications in-place, contributing to the efficiency of the solution.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Iterate through the characters of the input string using a loop.\n2. Check if the current character is the same as the previous one or adjacent in the alphabet.\n3. If the condition is met, increment the counter and skip the next character to avoid double counting.\n4. Continue this process until the end of the string is reached.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n/**\n * The function removes almost-equal characters from the given string.\n * Two characters are almost-equal if they are the same or adjacent in the alphabet.\n * @param word The input string\n * @return The minimum number of operations needed to remove almost-equal characters\n */\nclass Solution {\n    public int removeAlmostEqualCharacters(String word) {\n        // Initialize a counter for the number of operations\n        int count = 0;\n        \n        // Create a StringBuilder from the input string\n        StringBuilder sb = new StringBuilder(word);\n        \n        // Iterate through the characters in the string\n        for (int i = 1; i < word.length(); i++) {\n            // Check if the current character is the same as the previous one or adjacent in the alphabet\n            if (sb.charAt(i) == sb.charAt(i - 1) || Math.abs(sb.charAt(i) - sb.charAt(i - 1)) == 1) {\n                // If true, increment the counter and skip the next character to avoid double counting\n                count++;\n                i++;\n            }\n        }\n        \n        // Return the minimum number of operations needed\n        return count;\n    }\n}\n\n```
['Dynamic Programming', 'Greedy', 'Java']