Introduction

In this blog post, we will explore how to solve the LeetCode problem "3324" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, 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 target. Alice is going to type target on her computer using a special keyboard that has only two keys: Key 1 appends the character "a" to the string on the screen. Key 2 changes the last character of the string on the screen to its next character in the English alphabet. For example, "c" changes to "d" and "z" changes to "a". Note that initially there is an empty string "" on the screen, so she can only press key 1. Return a list of all strings that appear on the screen as Alice types target, in the order they appear, using the minimum key presses. Example 1: Input: target = "abc" Output: ["a","aa","ab","aba","abb","abc"] Explanation: The sequence of key presses done by Alice are: Press key 1, and the string on the screen becomes "a". Press key 1, and the string on the screen becomes "aa". Press key 2, and the string on the screen becomes "ab". Press key 1, and the string on the screen becomes "aba". Press key 2, and the string on the screen becomes "abb". Press key 2, and the string on the screen becomes "abc". Example 2: Input: target = "he" Output: ["a","b","c","d","e","f","g","h","ha","hb","hc","hd","he"] Constraints: 1 <= target.length <= 400 target consists only of lowercase English letters.

Explanation

Here's the breakdown of the optimal approach, complexity analysis, and the Python code:

• Approach: The solution iteratively builds the target string. At each position in the target string, it first appends 'a' characters until the length matches the current index + 1. Then, it changes the last character to the target character if it's different from the current last character, accumulating intermediate strings along the way.

• Complexity:

  • Runtime Complexity: O(N * M), where N is the length of the target string and M is the maximum difference between two consecutive characters.
  • Storage Complexity: O(N * L), where N is the length of target and L is average length of the strings added to the result.

Code

    def strings_on_screen(target):
    """
    Generates a list of strings that appear on the screen as Alice types the target,
    using the minimum number of key presses.

    Args:
        target: The target string.

    Returns:
        A list of strings that appear on the screen.
    """

    result = []
    current_string = ""

    for i in range(len(target)):
        # Append 'a' until the length is i + 1
        while len(current_string) < i + 1:
            current_string += "a"
            result.append(current_string)

        # Change the last character if needed
        while current_string[i] != target[i]:
            current_string = current_string[:i] + chr(((ord(current_string[i]) - ord('a') + 1) % 26) + ord('a'))
            result.append(current_string)

    return result