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2465. Shifting Letters Ii

Difficulty: Medium

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2465. 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.

Solution

class Solution {
    public String shiftingLetters(String s, int[][] shifts) {
        int n = s.length();
        int pref[] = new int[n];
        for (int current[] : shifts) {
            int u = current[0], v = current[1], dir = current[2];
            if (dir == 1) {
                pref[u]++;
                if (v + 1 < n) pref[v + 1]--;
            }
            else {
                pref[u]--;
                if (v + 1 < n) pref[v + 1]++;
            }
        }
        for (int i = 1; i < n; i++) pref[i] += pref[i - 1];
        StringBuilder res = new StringBuilder();
        for (int i = 0; i < n; i++) {
            char current = s.charAt(i);
            if (pref[i] == 0) {
                res.append(current);
                continue;
            }
            if (pref[i] < 0) {
                int time_back = Math.abs(pref[i]) % 26;
                while (time_back > 0) {
                    if (current == 'a') {
                        current = 'z';
                        time_back--;
                    }
                    else {
                        current--;
                        time_back--;
                    }
                }
                res.append(current);
            }
            else {
                int time_forward = pref[i] % 26;
                while (time_forward > 0) {
                    if (current == 'z') {
                        current = 'a';
                        time_forward--;
                    }
                    else {
                        current++;
                        time_forward--;
                    }
                }
                res.append(current);
            }
        }
        return res.toString();
    }
}

Complexity Analysis

  • Time Complexity: O(?)
  • Space Complexity: O(?)

Approach

Detailed explanation of the approach will be added here