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').
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 <= 105s consists of lowercase English letters.shifts.length == s.length0 <= shifts[i] <= 109public 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();
}
}