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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. 1 <= X <= 10^9
2. 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.

• Java
• C++
• Python
• Go

• 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/broken-calculator/
  // 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
  }
  

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