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Letter Combinations of a Phone Number

This question is asked by Google. Given a string of digits, return all possible text messages those digits could send.

Note: The mapping of digits to letters is as follows:

  • 0 → null
  • 1 → null
  • 2 → "abc"
  • 3 → "def"
  • 4 → "ghi"
  • 5 → "jkl"
  • 6 → "mno"
  • 7 → "pqrs"
  • 8 → "tuv"
  • 9 → "wxyz"

Example(s)​

Example 1:

Input: digits = "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]
Explanation:
Digit 2 maps to "abc"
Digit 3 maps to "def"
All combinations: a+d, a+e, a+f, b+d, b+e, b+f, c+d, c+e, c+f
Total: 3 × 3 = 9 combinations

Example 2:

Input: digits = "2"
Output: ["a", "b", "c"]
Explanation:
Digit 2 maps to "abc"
Only one digit, so all its letters.

Example 3:

Input: digits = ""
Output: []
Explanation:
Empty string, no combinations.

Example 4:

Input: digits = "79"
Output: ["pw", "px", "py", "pz", "qw", "qx", "qy", "qz", "rw", "rx", "ry", "rz", "sw", "sx", "sy", "sz"]
Explanation:
Digit 7 maps to "pqrs" (4 letters)
Digit 9 maps to "wxyz" (4 letters)
Total: 4 × 4 = 16 combinations

Solution​

The solution uses Backtracking (DFS):

  1. Digit mapping: Map each digit to its corresponding letters
  2. Backtracking: For each digit, try all possible letters
  3. Recursive building: Build combinations digit by digit
  4. Base case: When all digits processed, add combination to results

JavaScript Solution - Backtracking

/**
* Generate all letter combinations from phone number digits
* @param {string} digits - String of digits
* @return {string[]} - All possible letter combinations
*/
function letterCombinations(digits) {
// Handle empty input
if (!digits || digits.length === 0) {
  return [];
}

// Digit to letters mapping
const digitMap = {
  '0': '',
  '1': '',
  '2': 'abc',
  '3': 'def',
  '4': 'ghi',
  '5': 'jkl',
  '6': 'mno',
  '7': 'pqrs',
  '8': 'tuv',
  '9': 'wxyz'
};

const result = [];
const current = [];

/**
 * Backtracking function
 * @param {number} index - Current digit index
 */
function backtrack(index) {
  // Base case: processed all digits
  if (index === digits.length) {
    result.push(current.join(''));
    return;
  }
  
  // Get letters for current digit
  const digit = digits[index];
  const letters = digitMap[digit];
  
  // Skip if digit maps to empty string (0 or 1)
  if (letters.length === 0) {
    backtrack(index + 1);
    return;
  }
  
  // Try each letter for current digit
  for (const letter of letters) {
    current.push(letter);
    backtrack(index + 1);
    current.pop(); // Backtrack
  }
}

backtrack(0);
return result;
}

// Test cases
console.log('Example 1:', letterCombinations("23")); 
// ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]

console.log('Example 2:', letterCombinations("2")); 
// ["a", "b", "c"]

console.log('Example 3:', letterCombinations("")); 
// []

console.log('Example 4:', letterCombinations("79")); 
// ["pw", "px", "py", "pz", "qw", "qx", "qy", "qz", "rw", "rx", "ry", "rz", "sw", "sx", "sy", "sz"]
Output:
Click "Run Code" to execute the code and see the results.

Complexity​

  • Time Complexity: O(4^N × N) - Where N is the number of digits. In the worst case, each digit maps to 4 letters (7 or 9), so we have 4^N combinations, and each combination takes O(N) to build.
  • Space Complexity: O(N) - For the recursion stack and current combination (excluding output space).

Approach​

The solution uses Backtracking (DFS):

  1. Digit mapping:

    • Create a map from digits to their corresponding letters
    • Handle special cases: 0 and 1 map to empty strings

  2. Backtracking function:

    • backtrack(index): Generate combinations starting from digit at index
    • Process digits from left to right

  3. For each digit:

    • Get the letters mapped to the current digit
    • If empty (0 or 1), skip to next digit
    • Otherwise, try each letter for this digit

  4. Recursive building:

    • Add letter to current combination
    • Recursively process next digit
    • Backtrack by removing the letter

  5. Base case:

    • When index === digits.length, we've processed all digits
    • Add current combination to results

Key Insights​

  • Backtracking: Explore all possible letter combinations
  • Cartesian product: Each digit contributes one letter to the combination
  • Handle 0 and 1: These digits map to empty strings (skip them)
  • O(4^N) combinations: In worst case (all 7s or 9s)
  • Systematic exploration: Backtracking ensures all combinations are generated

Step-by-Step Example​

Let's trace through Example 1: digits = "23"

digits = "23"
digitMap = {
  '2': 'abc',
  '3': 'def'
}

result = []
current = []

backtrack(0):
  index = 0, digit = '2', letters = 'abc'
  
  For letter = 'a':
    current = ['a']
    backtrack(1):
      index = 1, digit = '3', letters = 'def'
      
      For letter = 'd':
        current = ['a', 'd']
        backtrack(2):
          index = 2 === length → add "ad" to result
        current.pop() → ['a']
      
      For letter = 'e':
        current = ['a', 'e']
        backtrack(2):
          index = 2 === length → add "ae" to result
        current.pop() → ['a']
      
      For letter = 'f':
        current = ['a', 'f']
        backtrack(2):
          index = 2 === length → add "af" to result
        current.pop() → ['a']
    
    current.pop() → []
  
  For letter = 'b':
    current = ['b']
    backtrack(1):
      ... (similar, generates "bd", "be", "bf")
    current.pop() → []
  
  For letter = 'c':
    current = ['c']
    backtrack(1):
      ... (similar, generates "cd", "ce", "cf")
    current.pop() → []

Result: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]

Visual Representation​

Example 1: digits = "23"

Digit 2: "abc"    Digit 3: "def"

Backtracking tree:
  backtrack(0)
    ├─ 'a' → backtrack(1)
    │   ├─ 'd' → "ad" ✓
    │   ├─ 'e' → "ae" ✓
    │   └─ 'f' → "af" ✓
    ├─ 'b' → backtrack(1)
    │   ├─ 'd' → "bd" ✓
    │   ├─ 'e' → "be" ✓
    │   └─ 'f' → "bf" ✓
    └─ 'c' → backtrack(1)
        ├─ 'd' → "cd" ✓
        ├─ 'e' → "ce" ✓
        └─ 'f' → "cf" ✓

Result: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]

Total combinations: 3 × 3 = 9

Edge Cases​

  • Empty string: Return [] (no combinations)
  • Single digit: Return all letters for that digit
  • Digits 0 or 1: Skip them (map to empty strings)
  • All same digit: Generate all combinations with that digit's letters
  • Mixed digits: Handle different letter counts per digit

Important Notes​

  • Digit mapping: Standard phone keypad mapping
  • 0 and 1: Map to empty strings (no letters)
  • Cartesian product: Each digit contributes one letter
  • O(4^N) combinations: In worst case
  • Backtracking: Systematically explores all combinations

Why Backtracking Works​

Key insight:

  • We need to generate all possible combinations of letters
  • Each digit contributes one letter to the combination
  • This is a Cartesian product of all letter sets
  • Backtracking systematically explores all possibilities

Optimal substructure:

  • To generate combinations for digits[index...], we:
    • Choose a letter for digits[index]
    • Recursively generate combinations for digits[index+1...]
  • This creates the recursive structure
  • Generate Parentheses: Similar backtracking problem
  • Combination Sum: Different backtracking problem
  • Permutations: Different combination problem
  • Subsets: Different combination problem

Takeaways​

  • Backtracking explores all letter combinations
  • Cartesian product of letter sets for each digit
  • Handle 0 and 1 as special cases (empty strings)
  • O(4^N) time exponential but necessary
  • Classic backtracking problem with practical applications