Leetcode 0319 Bulb Switcher

Problem statement

https://leetcode.com/problems/bulb-switcher/

Solution

Classical math brainteaser, where we need to check number of divisors for each number and bulb will be on in the end, if number of divisors is odd. Only squares have odd number of divisors, so the answer is int(sqrt(n)).

Complexity

It is O(1) for time and space

Code

class Solution:
    def bulbSwitch(self, n):
        return int(sqrt(n))