Last updated on July 19th, 2024 at 07:36 pm

Here, We see Letter Combinations of a Phone Number LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Topics

Backtracking, String

Companies

Amazon, Dropbox, Facebook, Google, Uber

Level of Question

Medium

Letter Combinations of a Phone Number LeetCode Solution

Problem Statement

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.

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”]

1. Letter Combinations of a Phone Number LeetCode Solution C++

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if (digits.empty()) return {};
        string phone_map[] = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        vector<string> output;
        backtrack("", digits, phone_map, output);
        return output;        
    }
private:
    void backtrack(string combination, string next_digits, string phone_map[], vector<string>& output) {
        if (next_digits.empty()) {
            output.push_back(combination);
        } else {
            string letters = phone_map[next_digits[0] - '2'];
            for (char letter : letters) {
                backtrack(combination + letter, next_digits.substr(1), phone_map, output);
            }
        }
    }
};

2. Letter Combinations of a Phone Number LeetCode Solution Java

class Solution {
    public List<String> letterCombinations(String digits) {
        if (digits.isEmpty()) return Collections.emptyList();
        String[] phone_map = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        List<String> output = new ArrayList<>();
        backtrack("", digits, phone_map, output);
        return output;
    }

    private void backtrack(String combination, String next_digits, String[] phone_map, List<String> output) {
        if (next_digits.isEmpty()) {
            output.add(combination);
        } else {
            String letters = phone_map[next_digits.charAt(0) - '2'];
            for (char letter : letters.toCharArray()) {
                backtrack(combination + letter, next_digits.substring(1), phone_map, output);
            }
        }
    }
}

3. Letter Combinations of a Phone Number Solution JavaScript

var letterCombinations = function(digits) {
    if (digits.length === 0) return [];
    const phone_map = ["abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"];
    const output = [];
    backtrack("", digits, phone_map, output);
    return output;
    function backtrack(combination, next_digits, phone_map, output) {
        if (next_digits.length === 0) {
            output.push(combination);
        } else {
            const letters = phone_map[next_digits[0] - '2'];
            for (const letter of letters) {
                backtrack(combination + letter, next_digits.slice(1), phone_map, output);
            }
        }
    }
};

4. Letter Combinations of a Phone Number Solution Python

class Solution(object):
    def letterCombinations(self, digits):
        if not digits:
            return []
        phone_map = {
            '2': 'abc',
            '3': 'def',
            '4': 'ghi',
            '5': 'jkl',
            '6': 'mno',
            '7': 'pqrs',
            '8': 'tuv',
            '9': 'wxyz'
        }
        def backtrack(combination, next_digits):
            if len(next_digits) == 0:
                output.append(combination)
            else:
                for letter in phone_map[next_digits[0]]:
                    backtrack(combination + letter, next_digits[1:])
        output = []
        backtrack("", digits)
        return output