Min Max Game

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

Apply the following algorithm on nums:

1. 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.
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]).
3. 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]).
4. Replace the array nums with newNums.
5. 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.

SOLUTION:

class Solution:
    def minMax(self, nums, i, j, minm):
        if j == i + 1:
            return nums[i]
        mid = (i + j) // 2
        a = self.minMax(nums, i, mid, True)
        b = self.minMax(nums, mid, j, False)
        if minm:
            return min(a, b)
        return max(a, b)

    def minMaxGame(self, nums: List[int]) -> int:
        n = len(nums)
        return self.minMax(nums, 0, n, True)