Leetcode: Bulb Switcher

Question

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.

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.
Hide Tags Math Brainteaser

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My solution

time complexity: O(n^2)

For $n^{th}$ bulb, check how many factors does n have?

class Solution(object):
    def bulbSwitch(self, n):
        """
        :type n: int
        :rtype: int
        """

        if n<=3:
            return n>=1

        res = 1
        for bulb_index in range(4,n+1):
            res = res+1 if self.helper(bulb_index)%2==0 else res    

        return res

    def helper(self,i):
        count = 0
        for ind in range(2,i+1):
            if i%ind==0:
                count += 1

        return count       

Error:
Time exceeds for input 9999

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Other’s Solution

time complexity: O(n)

$3 = 1\times3, 3\times1$
$4 = 1\times4, 2\times2, 4\times1$
$5 = 1\times5, 5\times1$
$6 = 1\times6, 2\times3, 3\times2, 6\times1$

the initial state of each bulb is off, if the number of factors (including 1) of n is even, bulb will be off eventually, otherwise it will be on. Actually, only when n is square number, the num of factors is odd.

class Solution(object):
    def bulbSwitch(self, n):
        """
        :type n: int
        :rtype: int
        """

        return int(math.sqrt(n))