# 1195-fizz-buzz-multithreaded Try it on leetcode ## Description

You have the four functions:

You are given an instance of the class FizzBuzz that has four functions: fizz, buzz, fizzbuzz and number. The same instance of FizzBuzz will be passed to four different threads:

Modify the given class to output the series [1, 2, "fizz", 4, "buzz", ...] where the ith token (1-indexed) of the series is:

Implement the FizzBuzz class:

 

Example 1:

Input: n = 15
Output: [1,2,"fizz",4,"buzz","fizz",7,8,"fizz","buzz",11,"fizz",13,14,"fizzbuzz"]

Example 2:

Input: n = 5
Output: [1,2,"fizz",4,"buzz"]

 

Constraints:

## Solution(Python) ```Python import threading class FizzBuzz(object): def __init__(self, n): self.__n = n self.__curr = 0 self.__cv = threading.Condition() # printFizz() outputs "fizz" def fizz(self, printFizz): """ :type printFizz: method :rtype: void """ for i in range(1, self.__n+1): with self.__cv: while self.__curr % 4 != 0: self.__cv.wait() self.__curr += 1 if i % 3 == 0 and i % 5 != 0: printFizz() self.__cv.notify_all() # printBuzz() outputs "buzz" def buzz(self, printBuzz): """ :type printBuzz: method :rtype: void """ for i in range(1, self.__n+1): with self.__cv: while self.__curr % 4 != 1: self.__cv.wait() self.__curr += 1 if i % 3 != 0 and i % 5 == 0: printBuzz() self.__cv.notify_all() # printFizzBuzz() outputs "fizzbuzz" def fizzbuzz(self, printFizzBuzz): """ :type printFizzBuzz: method :rtype: void """ for i in range(1, self.__n+1): with self.__cv: while self.__curr % 4 != 2: self.__cv.wait() self.__curr += 1 if i % 3 == 0 and i % 5 == 0: printFizzBuzz() self.__cv.notify_all() # printNumber(x) outputs "x", where x is an integer. def number(self, printNumber): """ :type printNumber: method :rtype: void """ for i in range(1, self.__n+1): with self.__cv: while self.__curr % 4 != 3: self.__cv.wait() self.__curr += 1 if i % 3 != 0 and i % 5 != 0: printNumber(i) self.__cv.notify_all() ```