2293 - Min Max Game.

Easy

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

• <code>1 <= numsi<= 10⁹</code>

• nums.length is a power of 2.