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# 991.Broken-Calculator

## 991. Broken Calculator

## 题目地址

<https://leetcode.com/problems/broken-calculator/>

## 题目描述

```
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
```

## 代码

### Approach #1 Work Backwards

Time `O(logY)` && Space `O(1)`

```java
class Solution {
  public int brokenCalc(int X, int Y) {
        int ans = 0;
    while (Y > X) {
      ans++;
      if (Y % 2 == 1) {
        Y++;
      } else {
        Y /= 2;
      }
    }

    return ans + X - Y;
  }
}
```