292. Nim Game
You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove the stones.
Both of you are very clever and have optimal strategies for the game. Write a function to determine whether you can win the game given the number of stones in the heap.
For example, if there are 4 stones in the heap, then you will never win the game: no matter 1, 2, or 3 stones you remove, the last stone will always be removed by your friend.
If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner?
For example, if the stone between 5 and 7. You can take 1 to 3 stone and then, your friend now can have to take the remaining 4 stones, then you must win. So if the number of the stone is not the multiple of 4, you can take 1 to 3 stone and lead your friend have to take from the stone, and you must win.
So, if you win, the number of the stone can not be the multiple of 4.
class Solution(object): def canWinNim(self, n): """ :type n: int :rtype: bool """ return not n % 4 == 0
class Solution(object): def canWinNim(self, n): """ :type n: int :rtype: bool """ return bool(n & 3)
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.
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.
Provided by @stefanpochmann
Because the light bulb is off, so when it turns on and off odd number of times, it can be on. And the divisors are pairs, like 1 and n, 2 and n/2, and so on. So, only square numbers have single divisors, because of the double divisor. So, we can only count out the square number.
class Solution(object): def bulbSwitch(self, n): """ :type n: int :rtype: int """ return int(math.sqrt(n))