1309. Decrypt String from Alphabet to Integer Mapping

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.

Solution:

class Solution:
    def freqAlphabets(self, s: str) -> str:
        
        i = len(s) - 1
        result = []
        
        while i >= 0:
            if s[i] == "#":
                temp = int(s[i - 1]) + int(s[i - 2]) * 10
                result.insert(0, chr(temp + 96))
                i -= 2
            else:
                result.insert(0, chr(int(s[i]) + 96))
                
            i -= 1
            
        return "".join(result)

It took me longer than it should have to figure out an approach for this problem. Eventually, I arrived at this solution. We iterate backwards through the string, and for each number we arrive at, we add the equivalent letter to the result. If we run into a # character, we take the following two-digit number and put its corresponding letter into our result.

Solution:

Try it on Leetcode