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Formatted question description: https://leetcode.ca/all/991.html
991. Broken Calculator
Level
Medium
Description
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
It is difficult to figure out how X
becomes Y
directly, but the problem can be solved in the opposite direction, which is try to make Y
become X
after some operations. To make Y
become X
, two operations are allowed. The first operation is half, which is divide Y
by 2, which is only allowed when Y
is even, and the second operation is increment, which is add 1 to Y
, which is allowed for all values of Y
.
When X > Y
or Y
is odd, obviously only increment can be used on Y
. When X < Y
and Y
is even, both half and increment can be used on Y
. Obviously using half costs fewer operations than using increment. Even if for the case where X * 2 > Y
it is better to use half first, since using half first and increments next costs 1 + X  Y / 2
operations, and using increments and half first (first increase Y
to X * 2
then divide Y
by 2) costs X * 2  Y + 1
, and obviously 1 + X  Y / 2 < X * 2  Y + 1
.

class Solution { public int brokenCalc(int X, int Y) { int count = 0; while (X < Y) { if (Y % 2 == 0) Y /= 2; else Y++; count++; } return count + X  Y; } }

// OJ: https://leetcode.com/problems/brokencalculator/ // Time: O(logY) // Space: O(1) class Solution { public: int brokenCalc(int X, int Y) { int ans = 0; while (Y > X) { ++ans; if (Y % 2) ++Y; else Y /= 2; } return ans + X  Y; } };

class Solution: def brokenCalc(self, startValue: int, target: int) > int: ans = 0 while startValue < target: if target & 1: target += 1 else: target >>= 1 ans += 1 ans += startValue  target return ans ############ class Solution(object): def brokenCalc(self, X, Y): """ :type X: int :type Y: int :rtype: 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

func brokenCalc(startValue int, target int) (ans int) { for startValue < target { if target&1 == 1 { target++ } else { target >>= 1 } ans++ } ans += startValue  target return }