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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1801_1900.s1844_replace_all_digits_with_characters;

// #Easy #String #2022_05_07_Time_1_ms_(70.12%)_Space_42.3_MB_(43.95%)

/**
 * 1844 - Replace All Digits with Characters\.
 *
 * Easy
 *
 * You are given a **0-indexed** string `s` that has lowercase English letters in its **even** indices and digits in its **odd** indices.
 *
 * There is a function `shift(c, x)`, where `c` is a character and `x` is a digit, that returns the x^th character after `c`.
 *
 * *   For example, `shift('a', 5) = 'f'` and `shift('x', 0) = 'x'`.
 *
 * For every **odd** index `i`, you want to replace the digit `s[i]` with `shift(s[i-1], s[i])`.
 *
 * Return `s` _after replacing all digits. It is **guaranteed** that_ `shift(s[i-1], s[i])` _will never exceed_ `'z'`.
 *
 * **Example 1:**
 *
 * **Input:** s = "a1c1e1"
 *
 * **Output:** "abcdef"
 *
 * **Explanation:** The digits are replaced as follows: 
 *
 * - s[1] -> shift('a',1) = 'b' 
 *
 * - s[3] -> shift('c',1) = 'd' 
 *
 * - s[5] -> shift('e',1) = 'f'
 *
 * **Example 2:**
 *
 * **Input:** s = "a1b2c3d4e"
 *
 * **Output:** "abbdcfdhe"
 *
 * **Explanation:** The digits are replaced as follows: 
 *
 * - s[1] -> shift('a',1) = 'b' 
 *
 * - s[3] -> shift('b',2) = 'd' 
 *
 * - s[5] -> shift('c',3) = 'f' 
 *
 * - s[7] -> shift('d',4) = 'h'
 *
 * **Constraints:**
 *
 * *   `1 <= s.length <= 100`
 * *   `s` consists only of lowercase English letters and digits.
 * *   `shift(s[i-1], s[i]) <= 'z'` for all **odd** indices `i`.
**/
public class Solution {
    public String replaceDigits(String s) {
        StringBuilder sb = new StringBuilder();
        for (char c : s.toCharArray()) {
            if (Character.isAlphabetic(c)) {
                sb.append(c);
            } else {
                sb.append((char) (sb.charAt(sb.length() - 1) + Character.getNumericValue(c)));
            }
        }
        return sb.toString();
    }
}