Min Steps to One

/*
Given a positive integer 'n', find and return the minimum number of steps that 'n' has to take to get reduced to 1. You can perform any one of the following 3 steps:
1.) Subtract 1 from it. (n = n - ­1) ,
2.) If its divisible by 2, divide by 2.( if n % 2 == 0, then n = n / 2 ) ,
3.) If its divisible by 3, divide by 3. (if n % 3 == 0, then n = n / 3 ).  
Write brute-force recursive solution for this.

Input format :
The first and the only line of input contains an integer value, 'n'.
Output format :
Print the minimum number of steps.

Constraints :
1 <= n <= 200
Time Limit: 1 sec

Sample Input 1 :
4
Sample Output 1 :
2 
Explanation of Sample Output 1 :
For n = 4
Step 1 :  n = 4 / 2  = 2
Step 2 : n = 2 / 2  =  1 

Sample Input 2 :
7
Sample Output 2 :
3
Explanation of Sample Output 2 :
For n = 7
Step 1 :  n = 7 ­- 1 = 6
Step 2 : n = 6  / 3 = 2 
Step 3 : n = 2 / 2 = 1  
*/
public class Solution {
    
	public static int countMinStepsToOne(int n) {
		//Your code goes here
        if (n<=1)
            return 0;
        else if (n==2 || n==3)
            return 1;
        
       int output1=countMinStepsToOne(n-1);
       int output2=Integer.MAX_VALUE;
       int output3=Integer.MAX_VALUE;
        
       if (n%2==0)
           output2=countMinStepsToOne(n/2);
        
       if (n%3==0)
           output3=countMinStepsToOne(n/3);
        
       return Math.min(Math.min(output2,output3),output1)+1;
	}

}