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. 1 <= X <= 10^9
2. 1 <= Y <= 10^9

Related Topics:
Math, Greedy

Similar Questions:

• 2 Keys Keyboard (Medium)

// OJ: https://leetcode.com/problems/broken-calculator/
// Author: github.com/lzl124631x
// Time: O(logY)
// Space: O(1)
class Solution {
public:
    int brokenCalc(int X, int Y) {
        int ans = 0;
        while (Y > X) {
            ++ans;
            if (Y % 2) ++Y;
            else Y /= 2;
        }
        return ans + X - Y;
    }
};