Bulb Switcher

Problem

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. Find how many bulbs are on after n rounds.

Example:

Given n = 3. 

At first, the three bulbs are [off, off, off].
After first round, the three bulbs are [on, on, on].
After second round, the three bulbs are [on, off, on].
After third round, the three bulbs are [on, off, off]. 

So you should return 1, because there is only one bulb is on.

Solution

public class Solution {
    public int bulbSwitch(int n) {
        return (int) Math.sqrt(n);
    }
}

Analysis

A bulb is on iff when it's toggled odd number of times
For each round i, the bulb who has divisor i will be toggled i, 2i, 3i, etc..
Therefore, as long as the bulb has odd number of divisors, it will be on at the end
We know that divisors always come as an pair 6 = 2 * 3, 8 = 2 * 4
Except that, for square number, there is an extra divisor sqrt(n)
Hence, square number always has odd number of divisors
To count how many square number we have until n, we just need to calculate sqrt(n)
Cause all the square number come as 1^2, 2^2, 3^2, ... sqrt(n)^2