Introduction

In this blog post, we will explore how to solve the LeetCode problem "3307" 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

Alice and Bob are playing a game. Initially, Alice has a string word = "a". You are given a positive integer k. You are also given an integer array operations, where operations[i] represents the type of the ith operation. Now Bob will ask Alice to perform all operations in sequence: If operations[i] == 0, append a copy of word to itself. If operations[i] == 1, 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 performing all the operations. Note that the character 'z' can be changed to 'a' in the second type of operation. Example 1: Input: k = 5, operations = [0,0,0] Output: "a" Explanation: Initially, word == "a". Alice performs the three operations as follows: Appends "a" to "a", word becomes "aa". Appends "aa" to "aa", word becomes "aaaa". Appends "aaaa" to "aaaa", word becomes "aaaaaaaa". Example 2: Input: k = 10, operations = [0,1,0,1] Output: "b" Explanation: Initially, word == "a". Alice performs the four operations as follows: Appends "a" to "a", word becomes "aa". Appends "bb" to "aa", word becomes "aabb". Appends "aabb" to "aabb", word becomes "aabbaabb". Appends "bbccbbcc" to "aabbaabb", word becomes "aabbaabbbbccbbcc". Constraints: 1 <= k <= 1014 1 <= operations.length <= 100 operations[i] is either 0 or 1. The input is generated such that word has at least k characters after all operations.

Explanation

Here's a breakdown of the solution:

• Work Backwards: Instead of building the final string, we can efficiently determine the kth character by reversing the operations. This avoids needing to construct extremely long strings.
• Track Length: Keep track of the current length of the string. When an operation of type 0 is encountered in reverse, halve the index k (rounding up or down appropriately) since the string was duplicated. When an operation of type 1 is encountered in reverse, decrement the character (modulo 26).

• Early Exit: If k becomes 1, we know we're looking at the very first character. We can then simply compute how many times 'a' has been incremented along the way.

• Runtime Complexity: O(N), where N is the length of the operations array.

• Storage Complexity: O(1).

Code

    def find_kth_character(k: int, operations: list[int]) -> str:
    """
    Finds the value of the kth character in word after performing all the operations.
    """

    k = int(k)  # Ensure k is an integer
    shift = 0
    length = 1  # Initial length of the string "a"

    for op in operations:
        if op == 0:
            length *= 2
        else:
            length *= 2
            shift = (shift + length // 2) % 26

    for op in reversed(operations):
        if op == 0:
            length //= 2
            if k > length:
                k -= length
        else:
            length //= 2
            if k > length:
                shift = (shift - length) % 26
                k -= length

    return chr(ord('a') + shift % 26)