17. Letter Combinations of a Phone Number

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

This is a classic recursion problem.

Solution:

• Use an array to store mapping of digits to Strings as seen in the telephone keypad. The mapping would be {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}
• If the input String is a digit (2 <= digit <= 9), then we would add each character in the mapped String to the output and return the letter combinations (e.g the letter combinations of 2 are {"a", "b", "c"}
• If there is more than 1 digit in the input, then we would apply the following steps
• Get mapped String "digitLetters" for first character
• Get letter combinations for remaining substring (starting from the 2nd character in input)
• For each letter combination in the combinations for the remaining substring
• For each character c in digitLetters, concatenate c and letter combination
• Add the resulting new letter combination to the output list 

public class Solution {
  String[] digitMap = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
  
  public List<String> letterCombinations(String digits) {
    List<String> letterCombo = new ArrayList<String>();
    if (digits.length() == 0) {
      return letterCombo;
    }
    
    String digitLetters = digitMap[digits.charAt(0) - '0'];
    if (digits.length() == 1) {
      for (int i = 0; i < digitLetters.length(); i++) {
        letterCombo.add(String.valueOf(digitLetters.charAt(i)));
      }
      return letterCombo;
    } else {
      List<String> subLetterCombo = letterCombinations(digits.substring(1));
      for (String s : subLetterCombo) {
        for (int i = 0;i < digitLetters.length(); i++) {
          letterCombo.add(digitLetters.charAt(i) + s);
        }
      }
      return letterCombo;
    }
  }
}