Minimum moves to reach target score

We should use double action as late as possible, as many times as possible. We can use a greedy approach to solve this problem.

class Solution:
    def minMoves(self, target: int, maxDoubles: int) -> int:
        res = 0
        while target > 1 and maxDoubles > 0:
            res += 1 + target % 2
            maxDoubles -= 1
            target //= 2
        return target - 1 + res

Time complexity: O(log(n)), n is the target.
Space complexity: O(1)