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Formatted question description: https://leetcode.ca/all/2381.html

2381. Shifting Letters II

  • Difficulty: Medium.
  • Related Topics: Array, String, Prefix Sum.
  • Similar Questions: The Skyline Problem, Range Sum Query - Mutable, Range Addition, Shifting Letters, Maximum Population Year, Describe the Painting.

Problem

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.

Solution (Java, C++, Python)

  • class Solution {
    public String shiftingLetters(String s, int[][] shifts) {
        int[] diff = new int[s.length() + 1];
        int l;
        int r;
        for (int[] shift : shifts) {
            l = shift[0];
            r = shift[1] + 1;
            diff[l] += 26;
            diff[r] += 26;
            if (shift[2] == 0) {
                diff[l]--;
                diff[r]++;
            } else {
                diff[l]++;
                diff[r]--;
            }
            diff[l] %= 26;
            diff[r] %= 26;
        }
        StringBuilder sb = new StringBuilder();
        int current = 0;
        int val;
        for (int i = 0; i < s.length(); ++i) {
            current += diff[i];
            val = s.charAt(i) - 'a';
            val += current;
            val %= 26;
            sb.append((char) ('a' + val));
        }
        return sb.toString();
    }
}

############

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:
    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)
        )

############

# 2381. Shifting Letters II
# https://leetcode.com/problems/shifting-letters-ii/

class Solution:
    def shiftingLetters(self, s: str, shifts: List[List[int]]) -> str:
        n = len(s)
        A = [ord(x) - ord('a') for x in s]
        B = [0] * (n + 1)
        
        for start, end, direction in shifts:
            c = 1 if direction == 1 else -1
            
            B[start] += c
            B[end + 1] -= c

        for i in range(1, len(B)):
            B[i] += B[i - 1]

        for index, (a, b) in enumerate(zip(A, B)):
            A[index] += b
            A[index] %= 26
            
        return "".join([chr(x + ord("a")) for x in A])


  • 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)
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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