# 848. Shifting Letters

## Description

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"
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

## Solutions

• class Solution {
public String shiftingLetters(String s, int[] shifts) {
char[] cs = s.toCharArray();
int n = cs.length;
long t = 0;
for (int i = n - 1; i >= 0; --i) {
t += shifts[i];
int j = (int) ((cs[i] - 'a' + t) % 26);
cs[i] = (char) ('a' + j);
}
return String.valueOf(cs);
}
}

• class Solution {
public:
string shiftingLetters(string s, vector<int>& shifts) {
long long t = 0;
int n = s.size();
for (int i = n - 1; ~i; --i) {
t += shifts[i];
int j = (s[i] - 'a' + t) % 26;
s[i] = 'a' + j;
}
return s;
}
};

• class Solution:
def shiftingLetters(self, s: str, shifts: List[int]) -> str:
n, t = len(s), 0
s = list(s)
for i in range(n - 1, -1, -1):
t += shifts[i]
j = (ord(s[i]) - ord('a') + t) % 26
s[i] = ascii_lowercase[j]
return ''.join(s)


• func shiftingLetters(s string, shifts []int) string {
t := 0
n := len(s)
cs := []byte(s)
for i := n - 1; i >= 0; i-- {
t += shifts[i]
j := (int(cs[i]-'a') + t) % 26
cs[i] = byte('a' + j)
}
return string(cs)
}