[LeetCode February Challange] Day 21 - Broken Calculator

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

---

Solution

Time complexity : O(logY)
Space complexity : O(1)

class Solution {
public:
    int brokenCalc(int X, int Y) {
        int ans = 0;
        while (X < Y) {
            Y = Y % 2 ? Y + 1 : Y / 2;
            ++ans;
        }
        return ans + (X - Y);
    }
};

X 至 Y 沒什麼線索可以思考，由 Y 至 X 的話…

• 若 Y 是奇數的話，一定是 某階段的 X - 1 來的。
• 若 Y 是偶數的話，一定是 某階段的 X * 2 來的。

若 Y < X，X 只能一直 -1 直到 Y。