Decrypt String from Alphabet to Integer Mapping

Problem

You are given a string s formed by digits 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.

The test cases are generated so 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"

Constraints:

1 <= s.length <= 1000
s consists of digits and the '#' letter.
s will be a valid string such that mapping is always possible.

Solution (Java)

class Solution {
    public String freqAlphabets(String s) {
        Map<String, String> map = new HashMap<>();
        map.put("1", "a");
        map.put("2", "b");
        map.put("3", "c");
        map.put("4", "d");
        map.put("5", "e");
        map.put("6", "f");
        map.put("7", "g");
        map.put("8", "h");
        map.put("9", "i");
        map.put("10#", "j");
        map.put("11#", "k");
        map.put("12#", "l");
        map.put("13#", "m");
        map.put("14#", "n");
        map.put("15#", "o");
        map.put("16#", "p");
        map.put("17#", "q");
        map.put("18#", "r");
        map.put("19#", "s");
        map.put("20#", "t");
        map.put("21#", "u");
        map.put("22#", "v");
        map.put("23#", "w");
        map.put("24#", "x");
        map.put("25#", "y");
        map.put("26#", "z");
        StringBuilder sb = new StringBuilder();
        int i = 0;
        while (i < s.length()) {
            if ((Integer.parseInt("" + s.charAt(i)) == 1 || Integer.parseInt("" + s.charAt(i)) == 2)
                    && i + 1 < s.length()
                    && i + 2 < s.length()
                    && s.charAt(i + 2) == '#') {
                sb.append(map.get(s.substring(i, i + 3)));
                i += 3;
            } else {
                sb.append(map.get("" + s.charAt(i)));
                i++;
            }
        }
        return sb.toString();
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).