# Solving Leetcode Interviews in Seconds with AI: Decrypt String from Alphabet to Integer Mapping


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1309" 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
	> You are given a string s formed by digits and '#'. We want to map s to English lowercase characters as follows:  Characters ('a' to 'i') are represented by ('1' to '9') respectively. Characters ('j' to 'z') are represented by ('10#' to '26#') respectively.  Return the string formed after mapping. The test cases are generated so that a unique mapping will always exist.   Example 1:  Input: s = "10#11#12" Output: "jkab" Explanation: "j" -> "10#" , "k" -> "11#" , "a" -> "1" , "b" -> "2".  Example 2:  Input: s = "1326#" Output: "acz"    Constraints:  1 <= s.length <= 1000 s consists of digits and the '#' letter. s will be a valid string such that mapping is always possible.  

	# Explanation
	*   Iterate through the string `s` from right to left.
*   If we encounter a '#', we combine the preceding two digits to form a number between 10 and 26. Otherwise, we use the current digit directly.
*   Convert the number to the corresponding lowercase character and prepend it to the result string.

*   Runtime Complexity: O(n), where n is the length of the input string. Storage Complexity: O(1), excluding the space for the result string.

	
	# Code
	```python
	def freqAlphabets(s: str) -> str:
    """
    Maps a string of digits and '#' to lowercase characters.

    Args:
        s: The input string consisting of digits and '#'.

    Returns:
        The string formed after mapping.
    """

    result = ""
    i = len(s) - 1
    while i >= 0:
        if s[i] == '#':
            num = int(s[i - 2:i])
            result = chr(ord('a') + num - 1) + result
            i -= 3
        else:
            num = int(s[i])
            result = chr(ord('a') + num - 1) + result
            i -= 1
    return result
	```