17. Letter Combinations of a Phone Number

Each digit contributes one letter to the combination. Recurse position by position: for the current digit, try each of its letters, append it, recurse to the next digit, then undo. When all digits are placed, the built string is one combination.

Idea

A combination picks exactly one letter per digit. This is a tree of choices: at depth i we choose a letter for digits[i], then move to depth i+1. Backtracking walks this tree, so every root-to-leaf path is one valid combination.

Algorithm

1. Map each digit to its letters. If digits is empty, return [].
2. backtrack(index, path):
   • If index == len(digits), record path.
   • Otherwise, for each letter of digits[index], append it, recurse with index+1, then pop.

Worked example: "23"

• Digit 2 → a/b/c; digit 3 → d/e/f.
• Combinations: ad, ae, af, bd, be, bf, cd, ce, cf.

Complexity

• Time: O(4^n · n) — up to 4 letters per digit, n digits.
• Space: O(n) recursion depth (plus output).

import java.util.ArrayList;
import java.util.List;

class Solution {
    private static final String[] MAP = {
        "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"
    };

    public List<String> letterCombinations(String digits) {
        List<String> result = new ArrayList<>();
        if (digits.isEmpty()) {
            return result;
        }
        backtrack(digits, 0, new StringBuilder(), result);
        return result;
    }

    private void backtrack(String digits, int index, StringBuilder path, List<String> result) {
        if (index == digits.length()) {
            result.add(path.toString());
            return;
        }
        String letters = MAP[digits.charAt(index) - '0'];
        for (int i = 0; i < letters.length(); i++) {
            path.append(letters.charAt(i));
            backtrack(digits, index + 1, path, result);
            path.deleteCharAt(path.length() - 1);
        }
    }
}