""" Given a string s formed by digits ('0' - '9') and '#' . We want to map s to English lowercase characters as follows:

Characters ('a' to 'i') are represented by ('1' to '9') respectively.
Characters ('j' to 'z') are represented by ('10#' to '26#') respectively. 
Return the string formed after mapping.

It's guaranteed that a unique mapping will always exist.

 

Example 1:

Input: s = "10#11#12"
Output: "jkab"
Explanation: "j" -> "10#" , "k" -> "11#" , "a" -> "1" , "b" -> "2".
Example 2:

Input: s = "1326#"
Output: "acz"
Example 3:

Input: s = "25#"
Output: "y"
Example 4:

Input: s = "12345678910#11#12#13#14#15#16#17#18#19#20#21#22#23#24#25#26#"
Output: "abcdefghijklmnopqrstuvwxyz"
 

Constraints:

1 <= s.length <= 1000
s[i] only contains digits letters ('0'-'9') and '#' letter.
s will be valid string such that mapping is always possible. """
import re
class Solution:
    def freqAlphabets(self, s: str) -> str:
        ans = re.sub(r'(\d\d)#', lambda m: chr(ord('a') + int(m.group(1)) - 1), s, flags=re.I)
        return re.sub(r'(\d)', lambda m: chr(ord('a') + int(m.group(1)) - 1), ans)
    def freqAlphabets(self, s):
        return ''.join(chr(int(i[:2]) + 96) for i in re.findall(r'\d\d#|\d', s))
    def freqAlphabets(self, s):
        ans, i = [], 0
        while i < len(s):
            if i < len(s) - 2 and s[i + 2] == '#':
                ans.append(chr(int(s[i:i + 2]) + 96))
                i += 3
            else:
                ans.append(chr(int(s[i]) + 96))
                i += 1
        return "".join(ans)


print(Solution().freqAlphabets("12345678910#11#12#13#14#15#16#17#18#19#20#21#22#23#24#25#26#")) #Output: "abcdefghijklmnopqrstuvwxyz"