Introduction

In this blog post, we will explore how to solve the LeetCode problem "2549" 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 positive integer n, that is initially placed on a board. Every day, for 109 days, you perform the following procedure: For each number x present on the board, find all numbers 1 <= i <= n such that x % i == 1. Then, place those numbers on the board. Return the number of distinct integers present on the board after 109 days have elapsed. Note: Once a number is placed on the board, it will remain on it until the end. % stands for the modulo operation. For example, 14 % 3 is 2. Example 1: Input: n = 5 Output: 4 Explanation: Initially, 5 is present on the board. The next day, 2 and 4 will be added since 5 % 2 == 1 and 5 % 4 == 1. After that day, 3 will be added to the board because 4 % 3 == 1. At the end of a billion days, the distinct numbers on the board will be 2, 3, 4, and 5. Example 2: Input: n = 3 Output: 2 Explanation: Since 3 % 2 == 1, 2 will be added to the board. After a billion days, the only two distinct numbers on the board are 2 and 3. Constraints: 1 <= n <= 100

Explanation

Here's a breakdown of the solution and the code:

• High-Level Approach:

  • The core idea is to simulate the process until a fixed point is reached, i.e., no new numbers are added to the board. Since the number of days is extremely large (10⁹), direct simulation would be infeasible.
  • We maintain a set to keep track of the distinct numbers currently on the board.
  • We iterate, adding new numbers based on the modulo condition, until no further numbers are added in a given iteration.

• Complexity:

  • Runtime Complexity: O(n²) - In the worst case, we might need to iterate almost up to 'n' times (each adding a few elements), and the modulo operation takes O(n) each time. Since the board size is bound by 'n', and the number of iterations are also bound by n
  • Storage Complexity: O(n) - The maximum number of distinct elements on the board cannot exceed 'n'.

Code

    def distinct_integers(n: int) -> int:
    """
    Calculates the number of distinct integers present on the board after 10^9 days.

    Args:
        n: The initial integer on the board.

    Returns:
        The number of distinct integers present on the board.
    """

    board = {n}
    while True:
        new_numbers = set()
        for x in list(board):  # Iterate over a copy to avoid modification issues
            for i in range(1, n + 1):
                if x % i == 1:
                    new_numbers.add(i)

        original_board_size = len(board)
        board.update(new_numbers)
        if len(board) == original_board_size:
            break

    return len(board)