Introduction

In this blog post, we will explore how to solve the LeetCode problem "1415" 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

A happy string is a string that: consists only of letters of the set ['a', 'b', 'c']. s[i] != s[i + 1] for all values of i from 1 to s.length - 1 (string is 1-indexed). For example, strings "abc", "ac", "b" and "abcbabcbcb" are all happy strings and strings "aa", "baa" and "ababbc" are not happy strings. Given two integers n and k, consider a list of all happy strings of length n sorted in lexicographical order. Return the kth string of this list or return an empty string if there are less than k happy strings of length n. Example 1: Input: n = 1, k = 3 Output: "c" Explanation: The list ["a", "b", "c"] contains all happy strings of length 1. The third string is "c". Example 2: Input: n = 1, k = 4 Output: "" Explanation: There are only 3 happy strings of length 1. Example 3: Input: n = 3, k = 9 Output: "cab" Explanation: There are 12 different happy string of length 3 ["aba", "abc", "aca", "acb", "bab", "bac", "bca", "bcb", "cab", "cac", "cba", "cbc"]. You will find the 9th string = "cab" Constraints: 1 <= n <= 10 1 <= k <= 100

Explanation

Here's the approach to solve this problem:

• Calculate Total Happy Strings: Determine the total number of happy strings of length n. This can be done using the formula: 3 * 2^(n-1). If k is greater than this total, return an empty string.

• Lexicographical Generation: Generate the kth happy string lexicographically. Start with the first character, then recursively build the remaining string, ensuring that adjacent characters are different. Calculate how many strings start with each character ('a', 'b', 'c') and adjust k accordingly to determine the correct starting character. Repeat this process for the subsequent characters.

• Base case and pruning: If n = 0, we have an empty string. To improve performance of the recursive function, we can compute and store intermediate results of the count of happy strings of a given length.

• Complexity:

  • Runtime Complexity: O(n) - The depth of the recursion is n.
  • Storage Complexity: O(n) - The depth of the recursion and the happy string being constructed.

Code

    def getHappyString(n: int, k: int) -> str:
    """
    Finds the kth happy string of length n.

    Args:
        n: The length of the happy string.
        k: The index of the happy string to find (1-indexed).

    Returns:
        The kth happy string, or an empty string if it doesn't exist.
    """

    total_happy_strings = 3 * (2**(n - 1))
    if k > total_happy_strings:
        return ""

    result = ""

    def solve(curr_len, remaining_k, prev_char):
        nonlocal result
        if curr_len == n:
            return True

        chars = ['a', 'b', 'c']
        for char in chars:
            if char != prev_char:
                count = 2**(n - 1 - curr_len)
                if remaining_k <= count:
                    result += char
                    if solve(curr_len + 1, remaining_k, char):
                        return True
                    else:
                        return False  # Backtrack if needed

                else:
                    remaining_k -= count

        return False

    solve(0, k, '')
    return result