LeetCode Bulb Switcher 319, leetcodeswitcher

LeetCode Bulb Switcher 319, leetcodeswitcher

Change the light bulb color

There areNBulbs 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 ). forNTh round, you only toggle the last bulb. Find how many bulbs are on afterNRounds.

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.

Coding

public int bulbSwitch(int n){        int[] arrN = new int[n];        int count = 0;        for (int i = 0; i < arrN.length; i++) {            arrN[i] =0;        }        for (int i = 0; i < arrN.length; i++) {            for (int j = 0; j < arrN.length;j++ ) {                if(i==0){                    arrN[j] = 1;//                    j++;                }else{                    int k = i+1;                    if((j+1)%k==0){arrN[j]=(arrN[j]==1?0:1);}//                    j=j+i;                }            }        }                for (int i = 0; i < arrN.length; i++) {            if (arrN[i]==1) {                count++;            }        }        return count;    }

 

Running Timeout:

The color change of the light bulb only exists in its own factor.

1. prime number, only 1 and yourself.

2. Common number, which is paired with a factor.

3. the light bulb will not be extinguished only when the bulb is fully lit for an odd number of times.

This question is converted to the number of full counters !!!

For example, 3 --> 3

1 public static int bulbSwitch(int n){2     return (int)Math.sqrt(n);3 }