Write a Program to Find Min Max Game

# 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:

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.
```

/**
 * @param {number[]} nums
 * @return {number}
 */
var minMaxGame = function(nums) {
    let n = nums.length;
    if (n === 1) return nums[0];

    const out = []
    for (let i = 0; i < (n / 2); i++) {
        out[i] = i & 1 ? Math.max(nums[2*i], nums[2*i+1]) : Math.min(nums[2*i], nums[2*i+1]);
    }
    return minMaxGame(out);
};