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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2401_2500.s2451_odd_string_difference;

// #Easy #String #Hash_Table #Math #2022_12_15_Time_0_ms_(100.00%)_Space_40_MB_(91.61%)

/**
 * 2451 - Odd String Difference\.
 *
 * Easy
 *
 * You are given an array of equal-length strings `words`. Assume that the length of each string is `n`.
 *
 * Each string `words[i]` can be converted into a **difference integer array** `difference[i]` of length `n - 1` where `difference[i][j] = words[i][j+1] - words[i][j]` where `0 <= j <= n - 2`. Note that the difference between two letters is the difference between their **positions** in the alphabet i.e. the position of `'a'` is `0`, `'b'` is `1`, and `'z'` is `25`.
 *
 * *   For example, for the string `"acb"`, the difference integer array is `[2 - 0, 1 - 2] = [2, -1]`.
 *
 * All the strings in words have the same difference integer array, **except one**. You should find that string.
 *
 * Return _the string in_ `words` _that has different **difference integer array**._
 *
 * **Example 1:**
 *
 * **Input:** words = ["adc","wzy","abc"]
 *
 * **Output:** "abc"
 *
 * **Explanation:** 
 *
 * - The difference integer array of "adc" is [3 - 0, 2 - 3] = [3, -1].
 *
 * - The difference integer array of "wzy" is [25 - 22, 24 - 25]= [3, -1]. 
 *
 * - The difference integer array of "abc" is [1 - 0, 2 - 1] = [1, 1].
 *
 * The odd array out is [1, 1], so we return the corresponding string, "abc".
 *
 * **Example 2:**
 *
 * **Input:** words = ["aaa","bob","ccc","ddd"]
 *
 * **Output:** "bob"
 *
 * **Explanation:** All the integer arrays are [0, 0] except for "bob", which corresponds to [13, -13].
 *
 * **Constraints:**
 *
 * *   `3 <= words.length <= 100`
 * *   `n == words[i].length`
 * *   `2 <= n <= 20`
 * *   `words[i]` consists of lowercase English letters.
**/
public class Solution {
    public String oddString(String[] w) {
        int n = w[0].length() - 1;
        int[] x = new int[n];
        int s = 1;
        int y = 0;
        int index = 1;
        for (int i = 0; i < n; i++) {
            x[i] = w[0].charAt(i + 1) - w[0].charAt(i);
        }
        for (int i = 1; y * s == 0 || s + y < 3; i++) {
            boolean b = true;
            for (int j = 0; j < n; j++) {
                if (x[j] != w[i].charAt(j + 1) - w[i].charAt(j)) {
                    b = false;
                    break;
                }
            }
            if (b) {
                s++;
            } else {
                y++;
                index = i;
            }
        }
        return s == 1 ? w[0] : w[index];
    }
}