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2381. Shifting Letters II

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

  • 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;
    }
};

  • 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)
        )


  • 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)
}

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