[CI] 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 nth round, you only toggle the last bulb. Find how many bulbs are on after n rounds.

It’s quite a interesting problem and more like a math problem than a coding one. The direct solution is do the swith round by round and we can get the result at the final round. The time complexity of the direct solution is O(n2). The example code is as follows,

public int bulbSwitch(int n) {
        int[] lights = new int[n];
        int cnt = 0;
        int i = 0;
        // initailize the array
        for (i = 0; i < n; i++) {
            lights[i] = 0;
        }
        // do the round toggle
        for (i = 0; i < n; i++) {
            for (int j = i; j < n; j += i + 1) {
                lights[j] = lights[j] == 0 ? 1 : 0;
            }
        }
        // get the result
        for (i = 0; i < n; i++) {
            cnt += lights[i];
        }
        return cnt;
}

Seems like a perfect answer which definitely is not. When I submitted the above code on LeetCode OJ, the platform threw out an error of RUNNING TIME OUT. So it’s time to get the code optimized.
As you can see, each one of the bulb is turned on or off for n times which is the number of divisors. But still this cannot solve our problem for the time complexity is O(n2) too! What about printing the result to see whether there is a pattern? Let do it.
when n=40, the result is 1001000010000001000000001000000000010000 . And when n=60, the result is 100100001000000100000000100000000001000000000000100000000000 .
What do you see, guys? Yes, the mumber of each group which has only one 1 is an Arithmetic sequence. So we can get the final result without any toggling rounds. A half but passed solution is as follows,

public int bulbSwitch(int n) {
        int cnt = 0;
        for (int i = 3; n > 0; i += 2) {
            n -= i;
            cnt++;
        }
        return cnt;
}

That’s it.
The time complexity of the above code is O(n) and the code can pass the OJ’s test. But is this the most optimized one? No, of course. If you know more maths, the above code can be optimized to the time complexity of O(1) which means there is some formula to get the final result. Can you figure it out, buddy?
By the way, the Ultimate answer is  SQRT(n).
source page: https://leetcode.com/problems/bulb-switcher/