Comment on page


17. Letter Combinations of a Phone Number



Given a string containing digits from 2-9 inclusive, 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. Note that 1 does not map to any letters.
Input: "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
Although the above answer is in lexicographical order, your answer could be in any order you want.


Approach 1: Backtracking

Complexity Analysis
  • Time complexity : O(3^N×4^M) where N is the number of digits in the input that maps to 3 letters (e.g. 2, 3, 4, 5, 6, 8) and M is the number of digits in the input that maps to 4 letters (e.g. 7, 9), and N+M is the total number digits in the input.
  • Space complexity : O(3^N×4^M) since one has to keep 3^N×4^M solutions.
class Solution {
Map<String, String> phone = new HashMap<String, String>(){{
put("2", "abc");
put("3", "def");
put("4", "ghi");
put("5", "jkl");
put("6", "mno");
put("7", "pqrs");
put("8", "tuv");
put("9", "wxyz");
List<String> output = new ArrayList<String>();
public List<String> letterCombinations(String digits) {
if (digits.length() != 0) {
backtrack("", digits);
return output;
private void backtrack(String combination, String next_digits) {
if (next_digits.length() == 0) {
} else {
String digit = next_digits.substring(0, 1);
String letters = phone.get(digit);
for (int i = 0; i < letters.length(); i++) {
String letter = letters.substring(i, i + 1);
backtrack(combination + letter, next_digits.substring(1));