Remove Letter To Equalize Frequency

You are given a 0-indexed string word, consisting of lowercase English letters. You need to select one index and remove the letter at that index from word so that the frequency of every letter present in word is equal.

Return true if it is possible to remove one letter so that the frequency of all letters in word are equal, and false otherwise.

LeetCode Problem - 2423

class Solution {
    public boolean equalFrequency(String word) {
        // Create a map to store the frequency of each character in the input string
        Map<Character, Integer> frequencyMap = new HashMap<>();

        // Loop through each character in the word and update its frequency in the map
        for (char ch : word.toCharArray()) {
            frequencyMap.put(ch, frequencyMap.getOrDefault(ch, 0) + 1);
        }

        // Iterate through the characters in the frequency map
        for (char ch : frequencyMap.keySet()) {
            // Decrease the frequency of the current character by 1 to simulate its removal
            frequencyMap.put(ch, frequencyMap.get(ch) - 1);

            // Create a new map to count the frequencies of the remaining frequencies
            Map<Integer, Integer> freqCount = new HashMap<>();

            // Iterate through the values of the frequency map
            for (int freq : frequencyMap.values()) {
                // Only consider non-zero frequencies and update the frequency count map
                if (freq > 0) {
                    freqCount.put(freq, freqCount.getOrDefault(freq, 0) + 1);
                }
            }

            // If the frequency count map has only one unique frequency, return true
            if (freqCount.size() == 1) {
                return true;
            }

            // Restore the frequency of the current character after the check
            frequencyMap.put(ch, frequencyMap.get(ch) + 1);
        }

        // If no valid frequency condition is found, return false
        return false;
    }
}