"""
You are given a 0-indexed integer array nums whose length is a power of 2.

Apply the following algorithm on nums:

Let n be the length of nums. If n == 1, end the process. Otherwise, create a new 0-indexed integer array newNums of length n / 2.
For every even index i where 0 <= i < n / 2, assign the value of newNums[i] as min(nums[2 * i], nums[2 * i + 1]).
For every odd index i where 0 <= i < n / 2, assign the value of newNums[i] as max(nums[2 * i], nums[2 * i + 1]).
Replace the array nums with newNums.
Repeat the entire process starting from step 1.
Return the last number that remains in nums after applying the algorithm.



Example 1:


Input: nums = [1,3,5,2,4,8,2,2]
Output: 1
Explanation: The following arrays are the results of applying the algorithm repeatedly.
First: nums = [1,5,4,2]
Second: nums = [1,4]
Third: nums = [1]
1 is the last remaining number, so we return 1.
Example 2:

Input: nums = [3]
Output: 3
Explanation: 3 is already the last remaining number, so we return 3.


Constraints:

1 <= nums.length <= 1024
1 <= nums[i] <= 109
nums.length is a power of 2.
"""
from typing import List


class MinMaxGame:
    def minMaxGame(self, nums: List[int]) -> int:

        def dfs(arr, op):
            n = len(arr)
            if n == 1:
                return arr[0]
            if op == 0:
                return min(dfs(arr[:n // 2], 0), dfs(arr[n // 2:], 1))
            else:
                return max(dfs(arr[:n // 2], 0), dfs(arr[n // 2:], 1))

        return dfs(nums, 0)