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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2401_2500.s2423_remove_letter_to_equalize_frequency;

// #Easy #String #Hash_Table #Counting #2022_11_18_Time_1_ms_(75.41%)_Space_41.9_MB_(50.90%)

/**
 * 2423 - Remove Letter To Equalize Frequency\.
 *
 * Easy
 *
 * 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_.
 *
 * **Note:**
 *
 * *   The **frequency** of a letter `x` is the number of times it occurs in the string.
 * *   You **must** remove exactly one letter and cannot chose to do nothing.
 *
 * **Example 1:**
 *
 * **Input:** word = "abcc"
 *
 * **Output:** true
 *
 * **Explanation:** Select index 3 and delete it: word becomes "abc" and each character has a frequency of 1. 
 *
 * **Example 2:**
 *
 * **Input:** word = "aazz"
 *
 * **Output:** false
 *
 * **Explanation:** We must delete a character, so either the frequency of "a" is 1 and the frequency of "z" is 2, or vice versa. It is impossible to make all present letters have equal frequency. 
 *
 * **Constraints:**
 *
 * *   `2 <= word.length <= 100`
 * *   `word` consists of lowercase English letters only.
**/
@SuppressWarnings("java:S135")
public class Solution {
    private boolean isValidWord(int[] arr) {
        int temp = 0;
        for (int j = 0; j < 26; j++) {
            if (arr[j] == 0) {
                continue;
            }
            if (temp == 0) {
                temp = arr[j];
                continue;
            }
            if (arr[j] != temp) {
                return false;
            }
        }
        return true;
    }

    public boolean equalFrequency(String word) {
        boolean ans = false;
        // frequency array
        int[] arr = new int[26];
        for (int i = 0; i < word.length(); i++) {
            arr[word.charAt(i) - 'a'] += 1;
        }
        for (int i = 0; i < 26; i++) {
            if (arr[i] != 0) {
                arr[i] -= 1;
                if (isValidWord(arr)) {
                    ans = true;
                    break;
                }
                arr[i] += 1;
            }
        }
        return ans;
    }
}