Question

2381. Shifting Letters II

中文文档

Description

You are given a string s of lowercase English letters and a 2D integer array shifts where shifts[i] = [starti, endi, directioni]. For every i, shift the characters in s from the index starti to the index endi (inclusive) forward if directioni = 1, or shift the characters backward if directioni = 0.

Shifting a character forward means replacing it with the next letter in the alphabet (wrapping around so that 'z' becomes 'a'). Similarly, shifting a character backward means replacing it with the previous letter in the alphabet (wrapping around so that 'a' becomes 'z').

Return _the final string after all such shifts to _s_ are applied_.

 

Example 1:

Input: s = "abc", shifts = [[0,1,0],[1,2,1],[0,2,1]]
Output: "ace"
Explanation: Firstly, shift the characters from index 0 to index 1 backward. Now s = "zac".
Secondly, shift the characters from index 1 to index 2 forward. Now s = "zbd".
Finally, shift the characters from index 0 to index 2 forward. Now s = "ace".

Example 2:

Input: s = "dztz", shifts = [[0,0,0],[1,1,1]]
Output: "catz"
Explanation: Firstly, shift the characters from index 0 to index 0 backward. Now s = "cztz".
Finally, shift the characters from index 1 to index 1 forward. Now s = "catz".

 

Constraints:

• 1 <= s.length, shifts.length <= 5 * 104
• shifts[i].length == 3
• 0 <= starti <= endi < s.length
• 0 <= directioni <= 1
• s consists of lowercase English letters.

Solutions

Solution 1

class Solution:
  def shiftingLetters(self, s: str, shifts: List[List[int]]) -> str:
    n = len(s)
    d = [0] * (n + 1)
    for i, j, v in shifts:
      if v == 0:
        v = -1
      d[i] += v
      d[j + 1] -= v
    for i in range(1, n + 1):
      d[i] += d[i - 1]
    return ''.join(
      chr(ord('a') + (ord(s[i]) - ord('a') + d[i] + 26) % 26) for i in range(n)
    )

class Solution {
  public String shiftingLetters(String s, int[][] shifts) {
    int n = s.length();
    int[] d = new int[n + 1];
    for (int[] e : shifts) {
      if (e[2] == 0) {
        e[2]--;
      }
      d[e[0]] += e[2];
      d[e[1] + 1] -= e[2];
    }
    for (int i = 1; i <= n; ++i) {
      d[i] += d[i - 1];
    }
    StringBuilder ans = new StringBuilder();
    for (int i = 0; i < n; ++i) {
      int j = (s.charAt(i) - 'a' + d[i] % 26 + 26) % 26;
      ans.append((char) ('a' + j));
    }
    return ans.toString();
  }
}

class Solution {
public:
  string shiftingLetters(string s, vector<vector<int>>& shifts) {
    int n = s.size();
    vector<int> d(n + 1);
    for (auto& e : shifts) {
      if (e[2] == 0) {
        e[2]--;
      }
      d[e[0]] += e[2];
      d[e[1] + 1] -= e[2];
    }
    for (int i = 1; i <= n; ++i) {
      d[i] += d[i - 1];
    }
    string ans;
    for (int i = 0; i < n; ++i) {
      int j = (s[i] - 'a' + d[i] % 26 + 26) % 26;
      ans += ('a' + j);
    }
    return ans;
  }
};

func shiftingLetters(s string, shifts [][]int) string {
  n := len(s)
  d := make([]int, n+1)
  for _, e := range shifts {
    if e[2] == 0 {
      e[2]--
    }
    d[e[0]] += e[2]
    d[e[1]+1] -= e[2]
  }
  for i := 1; i <= n; i++ {
    d[i] += d[i-1]
  }
  ans := []byte{}
  for i, c := range s {
    j := (int(c-'a') + d[i]%26 + 26) % 26
    ans = append(ans, byte('a'+j))
  }
  return string(ans)
}