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LeetCode 991 Broken Calculator - Medium

991. Broken Calculator -- Medium

·1 min read·#LeetCode#Medium#Python

991. Broken Calculator — Medium

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Problem

  1. Broken Calculator -- Medium

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

Solution

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

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