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

Source: leetcode | Difficulty: MEDIUM_HARD
Tags: Mathematics

Greedy algorithms

Starter code: class Solution:
    def brokenCalc(self, X: int, Y: int) -> int:
        

Solutions:
class Solution:

	def brokenCalc(self, X: int, Y: int) -> int:
		res = 0
		while X < Y:
			res += Y % 2 + 1
			Y = int((Y + 1) / 2)
		return res + X - Y


class Solution:

	def brokenCalc(self, X: int, Y: int) -> int:
		result = 0
		while Y > X:
			result += 1
			if Y % 2:
				Y += 1
			else:
				Y //= 2
		return result + X - Y


class Solution:

	def brokenCalc(self, X: int, Y: int) -> int:
		count = 0
		while Y != X:
			if Y < X:
				count += X - Y
				Y = X
			elif Y % 2 == 0:
				Y //= 2
				count += 1
			else:
				Y += 1
				count += 1
		return count