codewars #35 Persistent Bugger. (6kyu)

 #35 Persistent Bugger.

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

import java.util.stream.Stream;
public class Persist {
	
	public static int persistence(long n) {
		// your code
		int count = 0;
		String temp = "";
		while((temp = String.valueOf(n)).length() > 1) {
			n = Stream.of(temp.split("")).mapToLong(Long::valueOf).reduce(1, (a, b) -> a * b);
			count++;
		}
		return count;
	}
}