Question

848. Shifting Letters

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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

Solution 1

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)

class Solution:
  def shiftingLetters(self, s: str, shifts: List[int]) -> str:
    n = len(s)
    d = [0] * (n + 1)
    for i, c in enumerate(s):
      v = ord(c) - ord('a')
      d[i] += v
      d[i + 1] -= v
    for i, x in enumerate(shifts):
      d[0] += x
      d[i + 1] -= x
    t = 0
    ans = []
    for i in range(n):
      d[i] %= 26
      ans.append(ascii_lowercase[d[i]])
      d[i + 1] += d[i]
    return ''.join(ans)

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;
  }
};

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)
}