1309. Decrypt String from Alphabet to Integer Mapping(Solution || Leetcode easy || Java)

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:

class Solution {
    public String freqAlphabets(String s) {
  //create a stringbuilder, so it will take the same space 
  //during modification 
      StringBuilder ans = new StringBuilder();
        
        int i = 0;
        
      while(i < s.length()){
          if(i < s.length()-2 && s.charAt(i+2) == '#'){
              ans.append((char)((Integer.valueOf(s.substring(i,i+2)))
                           + 'a' - 1));
              i+=3;
          }else{
              ans.append((char)((s.charAt(i) - '0') + 'a' - 1));
              i++;
          }
      }
        
        return ans.toString();
        }
        
    }

Runtime: 1 ms, faster than 97.12% of Java online submissions for Decrypt String from Alphabet to Integer Mapping.

Memory Usage: 40.3 MB, less than 96.49% of Java online submissions for Decrypt String from Alphabet to Integer Mapping.

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