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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"
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

Solutions

• Java
• C++
• Python
• Go

• 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)
  }
  

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