# Solving Leetcode Interviews in Seconds with AI: Find the K-th Character in String Game I


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3304" 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
	> Alice and Bob are playing a game. Initially, Alice has a string word = "a". You are given a positive integer k. Now Bob will ask Alice to perform the following operation forever:  Generate a new string by changing each character in word to its next character in the English alphabet, and append it to the original word.  For example, performing the operation on "c" generates "cd" and performing the operation on "zb" generates "zbac". Return the value of the kth character in word, after enough operations have been done for word to have at least k characters. Note that the character 'z' can be changed to 'a' in the operation.   Example 1:  Input: k = 5 Output: "b" Explanation: Initially, word = "a". We need to do the operation three times:  Generated string is "b", word becomes "ab". Generated string is "bc", word becomes "abbc". Generated string is "bccd", word becomes "abbcbccd".   Example 2:  Input: k = 10 Output: "c"    Constraints:  1 <= k <= 500  

	# Explanation
	Here's the breakdown of the solution and the Python code:

*   **High-Level Approach:**
    *   Simulate the string generation process until the string length is at least `k`.
    *   In each iteration, generate the new string and append it to the current string.
    *   Return the character at the `k-1` index of the final string.

*   **Complexity:**
    *   Runtime: O(k), Storage: O(k)

	
	# Code
	```python
	def solve():
    k = int(input())
    word = "a"
    while len(word) < k:
        new_word = ""
        for char in word:
            new_char = chr(((ord(char) - ord('a') + 1) % 26) + ord('a'))
            new_word += new_char
        word += new_word
    print(word[k-1])

solve()
	```