Introduction

In this blog post, we will explore how to solve the LeetCode problem "400" 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

Given an integer n, return the nth digit of the infinite integer sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...]. Example 1: Input: n = 3 Output: 3 Example 2: Input: n = 11 Output: 0 Explanation: The 11th digit of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... is a 0, which is part of the number 10. Constraints: 1 <= n <= 231 - 1

Explanation

Here's the breakdown of the solution:

• Determine the digit length: Find the length (number of digits) of the number that contains the nth digit. We iterate through lengths (1, 2, 3, ...) and calculate how many digits are occupied by numbers of that length.
• Identify the number: Once the length is determined, find the actual number that contains the nth digit. We calculate the starting number for the determined length and then add an offset to it based on the remaining n.

• Extract the digit: Finally, extract the specific digit from the number. Convert the number to a string and get the digit at the appropriate index.

• Runtime Complexity: O(log n). The loop iterates a number of times proportional to the number of digits in the final number.

• Storage Complexity: O(1). Constant extra space is used.

Code

    def findNthDigit(n: int) -> int:
    """
    Finds the nth digit in the infinite integer sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...].

    Args:
        n: The index of the digit to find.

    Returns:
        The nth digit of the sequence.
    """

    digit_length = 1
    count_of_numbers = 9
    while n > digit_length * count_of_numbers:
        n -= digit_length * count_of_numbers
        digit_length += 1
        count_of_numbers *= 10

    start_number = 10**(digit_length - 1)
    number = start_number + (n - 1) // digit_length
    index_within_number = (n - 1) % digit_length

    return int(str(number)[index_within_number])