Problem Description
You start with the integer 1 and want to reach a given target integer. In one move, you can either increment the current integer by 1 or double it. However, you can only use the double operation at most maxDoubles times. The goal is to determine the minimum number of moves required to reach the target.
Key Insights
- When doubling is available and the target is even, doubling (or working backwards by halving) is usually optimal.
- If the target is odd, it's best to subtract 1 (simulate the increment in reverse) to make it even.
- Once no doubling moves remain, the remaining distance must be covered with increments.
- Working backwards from the target to 1 simplifies the decision process, as it avoids overcounting moves.
Space and Time Complexity
Time Complexity: O(log(target)) in most cases, though in worst-case scenarios with many subtractions it can approach O(target) if doubling is exhausted quickly. Space Complexity: O(1)
Solution
We use a greedy approach working from the target backwards to 1. At each step, if we have remaining doubling moves and the target is even, we reverse a double (by halving the target) which corresponds to a double move in the forward process. If the target is odd, we subtract 1 to make it even, simulating an increment in reverse. If no doubling moves remain, we subtract 1 repeatedly until we reach 1, counting each subtraction as a move. This method ensures the number of moves is minimized.