Today is Sunday, I gonna review the tasks I've done this week, and finish today's leetcode's February LeetCoding Challenge with cpp.

February LeetCoding Challenge 21

Description

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. $1 \le X \le 10^9$
2. $1 \le Y \le 10^9$

Solution

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