There is a broken calculator that has the integer startValue
on its display initially. In one operation, you can:
- multiply the number on display by
2
, or - subtract
1
from the number on display.
Given two integers startValue
and target
, return the minimum number of operations needed to display target
on the calculator.
Example 1:
Input: startValue = 2, target = 3
Output: 2
Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.
Example 2:
Input: startValue = 5, target = 8
Output: 2
Explanation: Use decrement and then double {5 -> 4 -> 8}.
Example 3:
Input: startValue = 3, target = 10
Output: 3
Explanation: Use double, decrement and double {3 -> 6 -> 5 -> 10}.
Constraints:
-
1 <= startValue, target <= 109
SOLUTION:
class Solution:
def brokenCalc(self, startValue: int, target: int) -> int:
if startValue >= target:
return startValue - target
if target % 2 == 0:
return 1 + self.brokenCalc(startValue, target // 2)
else:
return 2 + self.brokenCalc(startValue, (target + 1) // 2)
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