Catch That Cow

 

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

 

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

 

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?Input

Line 1: Two space-separated integers: N and K

 

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

 

Sample Input

5 17Sample Output

4

 

Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

 

 

import java.util.*;
public class Main{
	static int a,b,num[],st,en;
	static A x[];
	public static void main(String args[]){
		Scanner reader=new Scanner(System.in);
		a=reader.nextInt();
		b=reader.nextInt();
        num=new int[200005];
        num[a]=1;
        A xx=new A(a,0);
        x=new A[200005];
        x[0]=xx;
        st=0;
        en=1;
        f();
        //System.out.println(num[b]);
	}
	public static void f(){
		for(;;){
		A xx=x[st];
		st++;
		//System.out.println(xx.step+" "+xx.k+" ");
		if(xx.step==b){
		    System.out.println(xx.k);
		    break;
		}
		if(xx.step>=1&&num[xx.step-1]==0){
			num[xx.step-1]=1;
			A y=new A(xx.step-1,xx.k+1);
			x[en]=y;
			en++;
		}
		if(xx.step<b&&num[xx.step+1]==0){
			num[xx.step+1]=1;
			A y=new A(xx.step+1,xx.k+1);
			x[en]=y;
			en++;
		}
		if(xx.step<b&&num[xx.step*2]==0){
			num[xx.step*2]=1;
			A y=new A(xx.step*2,xx.k+1);
			x[en]=y;
			en++;
		}
		}
	}
}

class A{
	int step;
	int k;
	A(int a,int b){
		step=a;
		k=b;
	}
}