LeetCode2273

class Solution {
    public List<String> removeAnagrams(String[] words) {
        // Convert the array into a list for easier element removal.
        List<String> res = new LinkedList<>(Arrays.asList(words));

        int index = 1;
        // Iterate through the list and check consecutive elements.
        // Continue as long as the list is not empty and the index is within range.
        while (!res.isEmpty() && index < res.size()) {
            // Check if the current word and the previous one are anagrams.
            if (isAnagram(res.get(index - 1), res.get(index))) {
                res.remove(index); // Remove the latter one if they are anagrams.
            } else {
                index++; // Move to the next element otherwise.
            }
        }
        return res;
    }

    private boolean isAnagram(String word1, String word2) {
        int[] arr = new int[26]; // Use an array to count character frequencies.

        // First, check if the lengths are different — if so, they can’t be anagrams.
        if (word1.length() != word2.length()) return false;

        // Count the frequency of each letter in word1.
        for (char c : word1.toCharArray()) {
            arr[c - 'a']++;
        }

        // Decrease the count using letters in word2.
        for (char c : word2.toCharArray()) {
            arr[c - 'a']--;
        }

        // If all counts are zero, they are anagrams.
        for (int count : arr) {
            if (count != 0) return false;
        }

        return true;
    }
}