2381

2381. Shifting Letters II (Medium)

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.

Companies: Veritas

Related Topics:
Array, String, Prefix Sum

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Solution 1. Difference Array

// OJ: https://leetcode.com/problems/shifting-letters-ii
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    string shiftingLetters(string s, vector<vector<int>>& A) {
        int N = s.size();
        vector<int> d(N);
        for (auto &shift : A) {
            int start = shift[0], end = shift[1], direction = shift[2], diff = direction == 0 ? -1 : 1;
            d[start] = (d[start] + diff + 26) % 26;
            if (end + 1 < N) d[end + 1] = (d[end + 1] - diff + 26) % 26;
        }
        for (int i = 0, diff = 0; i < N; ++i) {
            diff = (diff + d[i] + 26) % 26;
            s[i] = 'a' + (s[i] - 'a' + diff) % 26;
        }
        return s;
    }
};