Keypad Permutation

Problem

Phone has letters on the number keys. for example, number 2 has ABC on it, number 3 has DEF, 4 number has GHI,... , and number 9 has WXYZ. Write a program that will print out all of the possible combination of those letters depending on the input.  

Solution

HashMap

public static ArrayList<String> letterCombinations(String digits) {
  StringBuilder sb = new StringBuilder();
  ArrayList<String> res = new ArrayList<String>();
  
    if(digits == null) {
        return res;
    }
    
    //hashmap to store the key pad info
    Map<Character, char[]> hm = new HashMap<Character, char[]>();
    hm.put('2', new char[]{'a', 'b', 'c'});
    hm.put('3', new char[]{'d', 'e', 'f'});
    hm.put('4', new char[]{'g', 'h', 'i'});
    hm.put('5', new char[]{'j', 'k', 'l'});
    hm.put('6', new char[]{'m', 'n', 'o'});
    hm.put('7', new char[]{'p', 'q', 'r', 's'});
    hm.put('8', new char[]{'t', 'u', 'v'});
    hm.put('9', new char[]{'w', 'x', 'y', 'z'});
    
    helper(digits, res, sb, hm, 0);
    
    return res;

}

public static void helper(String digits, ArrayList<String> res, StringBuilder sb, Map<Character, char[]> hm, int pos) {
  if(pos == digits.length()) {
      res.add(sb.toString());
      return;
  }
  
  for(int i=0; i<hm.get(digits.charAt(pos)).length; i++) {
      sb.append(hm.get(digits.charAt(pos))[i]);
      helper(digits, res, sb, hm, pos+1);
      sb.deleteCharAt(sb.length()-1);
  }
}