// Source : https://leetcode.com/problems/broken-calculator/
// Author : Hao Chen
// Date   : 2019-05-01

/***************************************************************************************************** 
 *
 * On a broken calculator that has a number showing on its display, we can perform two operations:
 * 
 * 	Double: Multiply the number on the display by 2, or;
 * 	Decrement: Subtract 1 from the number on the display.
 * 
 * Initially, the calculator is displaying the number X.
 * 
 * Return the minimum number of operations needed to display the number Y.
 * 
 * Example 1:
 * 
 * Input: X = 2, Y = 3
 * Output: 2
 * Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.
 * 
 * Example 2:
 * 
 * Input: X = 5, Y = 8
 * Output: 2
 * Explanation: Use decrement and then double {5 -> 4 -> 8}.
 * 
 * Example 3:
 * 
 * Input: X = 3, Y = 10
 * Output: 3
 * Explanation:  Use double, decrement and double {3 -> 6 -> 5 -> 10}.
 * 
 * Example 4:
 * 
 * Input: X = 1024, Y = 1
 * Output: 1023
 * Explanation: Use decrement operations 1023 times.
 * 
 * Note:
 * 
 * 	1 <= X <= 10^9
 * 	1 <= Y <= 10^9
 ******************************************************************************************************/


class Solution {
public:
    int brokenCalc(int X, int Y) {
        if (X >= Y) return X-Y ;
        if ( Y%2 ==0 ) return brokenCalc(X, Y/2) + 1;
        return brokenCalc(X, Y+1) + 1;
    }
};