[LeetCode] Replace All Digits with Characters

Leetcode: https://leetcode.com/problems/replace-all-digits-with-characters/

Problem:

You are given a 0-indexed string s that has lowercase English letters in its even indices and digits in its odd indices.

There is a function shift(c, x), where c is a character and x is a digit, that returns the xth character after c.

• For example, shift('a', 5) = 'f' and shift('x', 0) = 'x'.

For every odd index i, you want to replace the digit s[i] with shift(s[i-1], s[i]).

Return s after replacing all digits. It is guaranteed that shift(s[i-1], s[i]) will never exceed 'z'.

Example 1:

Input: s = "a1c1e1"
Output: "abcdef"
Explanation: The digits are replaced as follows:
- s[1] -> shift('a',1) = 'b'
- s[3] -> shift('c',1) = 'd'
- s[5] -> shift('e',1) = 'f'

Example 2:

Input: s = "a1b2c3d4e"
Output: "abbdcfdhe"
Explanation: The digits are replaced as follows:
- s[1] -> shift('a',1) = 'b'
- s[3] -> shift('b',2) = 'd'
- s[5] -> shift('c',3) = 'f'
- s[7] -> shift('d',4) = 'h'

Solution:

class Solution(object):
    def replaceDigits(self, s):
        """
        :type s: str
        :rtype: str
        """
        sentence = list(s)
        acc = 0
        for i in range(1, len(s), 2):
            sentence[i] = self.shift(sentence[acc],sentence[i])
            acc += 2
            
        return "".join(sentence)
    
    def shift(self, letter, number):
        return chr(ord(letter) + int(number))