solution.py

# Write a function, persistence, that takes in a positive parameter num and returns its multiplicative persistence,
# which is the number of times you must multiply the digits in num until you reach a single digit.

# For example:

# persistence(39) = > 3  # Because 3*9 = 27, 2*7 = 14, 1*4=4
# and 4 has only one digit.

# persistence(999) = > 4  # Because 9*9*9 = 729, 7*2*9 = 126,
# 1*2*6 = 12, and finally 1*2 = 2.

# persistence(4) = > 0  # Because 4 is already a one-digit number.
# persistence(39)  # returns 3, because 3*9=27, 2*7=14, 1*4=4
# and 4 has only one digit

# persistence(999)  # returns 4, because 9*9*9=729, 7*2*9=126,
# 1*2*6=12, and finally 1*2=2

# persistence(4)  # returns 0, because 4 is already a one-digit number
from functools import reduce


def persistence(n):
    n = list(str(n))

    count = 0
    while len(n) > 1:
        n = list(str(reduce(lambda a, b: int(a) * int(b), n)))
        count += 1

    return count