// 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; } };