2381. Shifting Letters II

Problem Link

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.

Solution

---

Python3

class BIT:
    def __init__(self, N):
        self.stree = [0] * (N + 2)
 
    def change(self, i, x):
        while i < len(self.stree):
            self.stree[i] += x
            i += i & (-i)
 
    def query(self, i):
        s = 0
 
        while i >= 1:
            s += self.stree[i]
            i -= i & (-i)
 
        return s
 
class Solution:
    def shiftingLetters(self, s: str, shifts: List[List[int]]) -> str:
        N = len(s)
        tree = BIT(N)
 
        for a, b, direction in shifts:
            d = 1 if direction == 1 else -1
 
            tree.change(a + 1, d)
            tree.change(b + 2, -d)
        
        res = []
        for i in range(N):
            k = (ord(s[i]) - ord('a') + tree.query(i + 1)) % 26
            res.append(chr(k + ord('a')))
 
        return "".join(res)