2293. 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.

class Solution {
public:
    int minMaxGame(vector<int>& nums) {
        while(nums.size() > 1) {
            vector<int> A;
            int n = nums.size();
            bool mi = true;
            for(int i = 0; i < n; i += 2, mi = !mi) {
                if(mi) {
                    A.push_back(min(nums[i], nums[i + 1]));
                } else {
                    A.push_back(max(nums[i], nums[i + 1]));
                }
            }
            swap(nums, A);
        }
        return nums[0];
    }
};