Minimum Steps to One

Question

Problem statement :

On a positive integer, 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 ).

Now the question is, given a positive integer n, find the minimum number of steps that takes n to 1

eg:

1. For n = 1 , output: 0
2. For n = 4 , output: 2 ( 4 /2 = 2 /2 = 1 )
3. For n = 7 , output: 3 ( 7 -1 = 6 /3 = 2 /2 = 1 )

I know the solution using dynamic programming and having a integer array. here is the code.

    public int bottomup(int n) {
            //here i am defining an integer array
            //Exception is thrown here, if the n values is high.
            public int[] bu = new int[n+1];
            bu[0] = 0;
            bu[1] = 0;
            for(int i=2;i<=n;i++) {
                    int r = 1+bu[i-1];
                    if(i%2 == 0) r = Math.min(r,1+bu[i/2]);
                    if(i%3 == 0) r = Math.min(r,1+bu[i/3]);
                    bu[i] = r;
            }
            return bu[n];
    }

But i want to solve this using less space.This solution throws OutofMemoryError in java if n=100000000.I don't want to increase my heap space.Does anyone has solution using less space?

Please note this problem cannot be solved using greedy algorthm.Using one while loop and check for divisible by 3 and divisible by 2 wont work.you have to use dynamic programming.please suggest if any has a solution using less space.

eg:

For n = 10 greedy algo is 10 /2 = 5 -1 = 4 /2 = 2 /2 = 1 which takes 4 steps.where as the solution should be 10-1 = 9 / 3 = 3 / 3 = 1, 3 steps.

I even tried topdown solution.

    public int[] td = null;
    public int topdown(int n) {
            if(n <= 1) return 0;
            int r = 1+topdown(n-1);
            if(td[n] == 0) {
                    if(n%2 == 0) r = Math.min(r,1+topdown(n/2));
                    if(n%3 == 0) r = Math.min(r,1+topdown(n/3));
                    td[n] = r;
            }
            return td[n];
    }

it is failing at n=10000.

Answer 1

One idea is that at any iteration you need the values only for r/3 to r. So you can keep discarding 1/3rd of the array.

I'm not familiar with Java, but with C++ you can use a double ended queue (deque):

You keep adding to the deque from the back.
When i = 6, you do not need bu[0] and bu[1]. So you pop out two elements from the front of the queue.

Random access [ ] is supported with deque container.

EDIT: Also as suggested in the comments, you should change your datatype to a smaller sized one since the maximum number of steps shall be of the order of ( (log N) to base 2)

EDIT2: As Dukeling pointed out, it seems that in Java there is no ready-made well-suited implementation for deque that would not compromise on time complexity. You can think of implementing it in your own way as C++ does (I heard it is implemented as a vector of vectors with the size of inner vectors being small as compared to the total number of elements).

Answer 2

Recursive Solution For this problem

public static int countMinStepsTo1(int n){

      if(n==1)
      {
          return 0;
      } 
      int count1,count2=Integer.MAX_VALUE,count3=Integer.MAX_VALUE;

      count1 = countMinStepsTo1(n-1);

       if((n%2)==0)
       {
           count2=countMinStepsTo1(n/2);
       }
        if((n%3)==0)
        {
            count3=countMinStepsTo1(n/3);
        }

        return 1+Math.min(count1,Math.min(count2,count3));
    }

DP Approch For this problem

public static int countstepsDP(int n)
    {
        int storage[] = new int[n+1];
        storage[1]=0;

        for(int i=2; i<=n;i++)
        {
            int min=storage[i-1];
            if(i%3==0)
            {
                if(min>storage[i/3])
                {
                    min=storage[i/3];
                }
            }
            if(i%2==0)
            {
                if(min>storage[i/2])
                {
                    min=storage[i/2];
                }
            }
            storage[i]=1+min;
        }

        return storage[n];

    }