# Solving Leetcode Interviews in Seconds with AI: Letter Combinations of a Phone Number


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "17" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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"]    Constraints:  0 <= digits.length <= 4 digits[i] is a digit in the range ['2', '9'].  

	# Explanation
	Here's a breakdown of the solution:

*   **Backtracking:** Use a recursive backtracking algorithm to explore all possible letter combinations.
*   **Digit-to-Letters Mapping:** Maintain a dictionary (or list) to map each digit to its corresponding letters.
*   **Base Case:** When the current combination's length equals the input digits' length, add it to the result.

*   **Runtime Complexity:** O(4<sup>N</sup>), where N is the length of the digits string.  **Storage Complexity:** O(N), for the recursion stack and to store temporary combinations.

	
	# Code
	```python
	def letterCombinations(digits: str) -> list[str]:
    """
    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.
    """

    digit_to_letters = {
        "2": "abc",
        "3": "def",
        "4": "ghi",
        "5": "jkl",
        "6": "mno",
        "7": "pqrs",
        "8": "tuv",
        "9": "wxyz",
    }

    result = []

    def backtrack(index: int, current_combination: str):
        if index == len(digits):
            if current_combination:  # Only add if combination is not empty
                result.append(current_combination)
            return

        digit = digits[index]
        letters = digit_to_letters[digit]

        for letter in letters:
            backtrack(index + 1, current_combination + letter)

    if not digits:
        return []

    backtrack(0, "")
    return result
	```