319. Bulb Switcher

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.

There are three types of numbers.

• prime number: only factor is 1 and itself, when at 1, the bulb is set, when at itself, the bulb is unset, and there is not other way the prime-number-index bulb get accessed again.
• regular number(not perfect square number), since each such number will have factors count in pairs. so the result of such bulb is reset to off.
• perfect square number, it may have several pairs of factors, but it also has a factor which x^2 = n, this x number only access the bulb once, so all the perfect-square number is will keep the bulb on.

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