implement-rand10-using-rand7.py

# V1 
# idea : 
# to make a random number as N * 10 form 
# then the  mod 10 outout  (output%10) is the desired answer.

# class Solution(object):
# 	import random
# 	def rand7():
# 		return random.randint(1, 7)
# 	def rand10(self):
# 		output = (rand7())**2
# 		if output > 40:
# 			output = output - 40 
# 		return output%10


# V2 
# https://blog.csdn.net/fuxuemingzhu/article/details/81809478

# The rand7() API is already defined for you.
# def rand7():
# @return a random integer in the range 1 to 7
class Solution:
    def rand10(self):
        """
        :rtype: int
        """
        return self.rand40() % 10 + 1
        
    def rand49(self):
        """
        random integer in 0 ~ 48
        """
        return 7 * (rand7() - 1) + rand7() - 1
        
    def rand40(self):
        """
        random integer in 0 ~ 40
        """
        num = self.rand49()
        while num >= 40:
            num = self.rand49()
        return num


# V3 
# Time:  O(1.189), counted by statistics, limit would be O(log10/log7) = O(1.183)
# Space: O(1)

import random
def rand7():
    return random.randint(1, 7)

# Reference: https://leetcode.com/problems/implement-rand10-using-rand7/discuss/151567/C++JavaPython-Average-1.199-Call-rand7-Per-rand10
class Solution(object):
    def __init__(self):
        self.__cache = []

    def rand10(self):
        """
        :rtype: int
        """
        def generate(cache):
            n = 32
            curr = sum((rand7()-1) * (7**i) for i in range(n))
            rang = 7**n
            while curr < rang//10*10:
                cache.append(curr%10+1)
                curr /= 10
                rang /= 10

        while not self.__cache:
            generate(self.__cache)
        return self.__cache.pop()

# V3' 
# Time:  O(2 * (1 + (9/49) + (9/49)^2 + ...)) = O(2/(1-(9/49)) = O(2.45)
# Space: O(1)
class Solution2(object):
    def rand10(self):
        """
        :rtype: int
        """
        while True:
            x = (rand7()-1)*7 + (rand7()-1)
            if x < 40:
                return x%10 + 1