README_EN.md

1309. Decrypt String from Alphabet to Integer Mapping

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Description

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.

Solutions

Python3

class Solution:
    def freqAlphabets(self, s: str) -> str:
        def get(s):
            return chr(ord('a') + int(s) - 1)

        i, n = 0, len(s)
        res = []
        while i < n:
            if i + 2 < n and s[i + 2] == '#':
                res.append(get(s[i: i + 2]))
                i += 3
            else:
                res.append(get(s[i]))
                i += 1
        return ''.join(res)

Java

class Solution {
    public String freqAlphabets(String s) {
        int i = 0, n = s.length();
        StringBuilder res = new StringBuilder();
        while (i < n) {
            if (i + 2 < n && s.charAt(i + 2) == '#') {
                res.append(get(s.substring(i, i + 2)));
                i += 3;
            } else {
                res.append(get(s.substring(i, i + 1)));
                i += 1;
            }
        }
        return res.toString();
    }

    private char get(String s) {
        return (char) ('a' + Integer.parseInt(s) - 1);
    }
}

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