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 &lt;= i &lt; n / 2, assign the value of newNums[i] as min(nums[2 * i], nums[2 * i + 1]).
	For every odd index i where 0 &lt;= i &lt; 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.

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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.


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Constraints:


	1 &lt;= nums.length &lt;= 1024
	1 &lt;= nums[i] &lt;= 109
	nums.length is a power of 2.


class Solution:
    def minMaxGame(self, a: List[int]) -> int:
        
        def solve(n):
            if n==1:
                return
            for i in range(n//2):
                if i%2:
                    a[i] = max (a[2*i], a[2*i+1])
                else:
                    a[i] = min (a[2*i], a[2*i+1])
            solve(n//2)
            return
        solve(len(a))
        return a[0]
class Solution {
    public int minMaxGame(int[] nums) {
        var isMin = true;
        var n=1;
        
        while(n<nums.length) {
            
            for(int i=0; i<nums.length; i+= n*2) {
                nums[i] = isMin ?
                    Math.min(nums[i], nums[i+n]) : 
                    Math.max(nums[i], nums[i+n]);
                isMin = !isMin;
            }
            
            n *= 2;
        }
        
        return nums[0];
        
    }
}
class Solution {
public:
    int minMaxGame(vector<int>& nums) {
        int n = nums.size(); 
        if(n==1) return nums[0];  // Base case
        vector<int> newNum(n/2); 
        for(int i=0; i<n/2; i++) { 
            if(i%2==0) newNum[i] = min(nums[2 * i], nums[2 * i + 1]); 
            else newNum[i] = max(nums[2 * i], nums[2 * i + 1]); 
        } 
        int res = minMaxGame(newNum); // Recursive call
        return res;
    }
};
var minMaxGame = function(nums) {
    while (nums.length > 1) {
        let half = nums.length / 2;
        for (let i = 0; i < half; i++)
            nums[i] = i % 2 === 0 ? Math.min(nums[2 * i], nums[2 * i + 1]) : Math.max(nums[2 * i], nums[2 * i + 1]);

        nums.length = half;
    }

    return nums[0];
};
Min Max Game