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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0801_0900.s0848_shifting_letters;

// #Medium #Array #String #2022_03_24_Time_11_ms_(88.11%)_Space_55.9_MB_(79.10%)

/**
 * 848 - Shifting Letters\.
 *
 * Medium
 *
 * You are given a string `s` of lowercase English letters and an integer array `shifts` of the same length.
 *
 * Call the `shift()` of a letter, the next letter in the alphabet, (wrapping around so that `'z'` becomes `'a'`).
 *
 * *   For example, `shift('a') = 'b'`, `shift('t') = 'u'`, and `shift('z') = 'a'`.
 *
 * Now for each `shifts[i] = x`, we want to shift the first `i + 1` letters of `s`, `x` times.
 *
 * Return _the final string after all such shifts to s are applied_.
 *
 * **Example 1:**
 *
 * **Input:** s = "abc", shifts = [3,5,9]
 *
 * **Output:** "rpl"
 *
 * **Explanation:** We start with "abc". 
 *
 * After shifting the first 1 letters of s by 3, we have "dbc". 
 *
 * After shifting the first 2 letters of s by 5, we have "igc". 
 *
 * After shifting the first 3 letters of s by 9, we have "rpl", the answer.
 *
 * **Example 2:**
 *
 * **Input:** s = "aaa", shifts = [1,2,3]
 *
 * **Output:** "gfd"
 *
 * **Constraints:**
 *
 * *   1 <= s.length <= 10⁵
 * *   `s` consists of lowercase English letters.
 * *   `shifts.length == s.length`
 * *   0 <= shifts[i] <= 10⁹
**/
public class Solution {
    public String shiftingLetters(String s, int[] shifts) {
        int n = shifts.length;
        int runningSum = 0;
        for (int i = n - 1; i >= 0; i--) {
            shifts[i] = (shifts[i] + runningSum) % 26;
            runningSum = shifts[i];
        }
        StringBuilder str = new StringBuilder();
        int i = 0;
        for (char c : s.toCharArray()) {
            int correctShift = (c - 'a' + shifts[i]) % 26;
            str.append((char) ('a' + correctShift));
            i++;
        }
        return str.toString();
    }
}