## DEV Community

Abhishek Chaudhary

Posted on

# Broken Calculator

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)
``````