Introduction

In this blog post, we will explore how to solve the LeetCode problem "319" 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

There are n bulbs that are initially off. You first turn on all the bulbs, then you turn off every second bulb. On the third round, you toggle every third bulb (turning on if it's off or turning off if it's on). For the ith round, you toggle every i bulb. For the nth round, you only toggle the last bulb. Return the number of bulbs that are on after n rounds. Example 1: Input: n = 3 Output: 1 Explanation: At first, the three bulbs are [off, off, off]. After the first round, the three bulbs are [on, on, on]. After the second round, the three bulbs are [on, off, on]. After the third round, the three bulbs are [on, off, off]. So you should return 1 because there is only one bulb is on. Example 2: Input: n = 0 Output: 0 Example 3: Input: n = 1 Output: 1 Constraints: 0 <= n <= 109

Explanation

Here's the breakdown of the solution:

• Perfect Squares: A bulb ends up on if it's toggled an odd number of times. The number of times a bulb is toggled corresponds to the number of its divisors. Numbers with an odd number of divisors are perfect squares.

• Counting Perfect Squares: The number of perfect squares less than or equal to n is the floor of the square root of n.

• Optimization: Directly calculate the square root and floor it to avoid simulating the toggling process.

• Runtime & Storage Complexity: O(1) runtime, O(1) storage.

Code

    import math

def bulb_switch(n: int) -> int:
    """
    Calculates the number of bulbs that are on after n rounds.

    Args:
        n: The number of bulbs.

    Returns:
        The number of bulbs that are on.
    """
    return int(math.sqrt(n))