848. Shifting Letters

Problem

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 <= 10^5

• s consists of lowercase English letters.

• shifts.length == s.length

• 0 <= shifts[i] <= 10^9

Solution (Java)

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();
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).