Letter Combinations of a Phone Number LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the  Letter Combinations of a Phone Number in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem –  Letter Combinations of a Phone Number– LeetCode Problem

 Letter Combinations of a Phone Number– LeetCode Problem

Problem:

Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent. Return the answer in any order.

A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.

200px Telephone keypad2.svg

Example 1:

Input: digits = "23"
Output: ["ad","ae","af","bd","be","bf","cd","ce","cf"]

Example 2:

Input: digits = ""
Output: []

Example 3:

Input: digits = "2"
Output: ["a","b","c"]

Constraints:

  • 0 <= digits.length <= 4
  • digits[i] is a digit in the range ['2', '9'].
Letter Combinations of a Phone Number– LeetCode Solutions
class Solution {
 public:
  vector<string> letterCombinations(string digits) {
    if (digits.empty())
      return {};

    vector<string> ans;

    dfs(digits, 0, "", ans);

    return ans;
  }

 private:
  const vector<string> digitToLetters{"",    "",    "abc",  "def", "ghi",
                                      "jkl", "mno", "pqrs", "tuv", "wxyz"};

  void dfs(const string& digits, int i, string&& path, vector<string>& ans) {
    if (i == digits.length()) {
      ans.push_back(path);
      return;
    }

    for (const char letter : digitToLetters[digits[i] - '0']) {
      path.push_back(letter);
      dfs(digits, i + 1, move(path), ans);
      path.pop_back();
    }
  }
};
class Solution {
  public List<String> letterCombinations(String digits) {
    if (digits.isEmpty())
      return new ArrayList<>();

    List<String> ans = new ArrayList<>();
    ans.add("");
    final String[] digitToLetters = {"",    "",    "abc",  "def", "ghi",
                                     "jkl", "mno", "pqrs", "tuv", "wxyz"};

    for (final char d : digits.toCharArray()) {
      List<String> temp = new ArrayList<>();
      for (final String s : ans)
        for (final char c : digitToLetters[d - '0'].toCharArray())
          temp.add(s + c);
      ans = temp;
    }

    return ans;
  }
}
class Solution:
  def letterCombinations(self, digits: str) -> List[str]:
    if not digits:
      return []

    ans = ['']
    digitToLetters = ['', '', 'abc', 'def', 'ghi',
                      'jkl', 'mno', 'pqrs', 'tuv', 'wxyz']

    for d in digits:
      temp = []
      for s in ans:
        for c in digitToLetters[ord(d) - ord('0')]:
          temp.append(s + c)
      ans = temp

    return

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