Description
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
1from 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
Solutions
Solution 1: Reverse Calculation
We can use a reverse calculation method, starting from target. If target is odd, then target = target + 1, otherwise target = target / 2. We accumulate the number of operations until target ≤ startValue. The final result is the number of operations plus startValue - target.
The time complexity is O(log n), where n is target. The space complexity is O(1).
PythonJavaC++GoTypeScript
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(code-box)
