2293. Min Max Game

Problem

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] <= 10^9

• nums.length is a power of 2.

Solution (Java)

class Solution {
    public int minMaxGame(int[] nums) {
        int n = nums.length;
        if (n == 1) {
            return nums[0];
        }
        int[] newNums = new int[n / 2];
        for (int i = 0; i < n / 2; i++) {
            if (i % 2 == 0) {
                newNums[i] = Math.min(nums[2 * i], nums[2 * i + 1]);
            } else {
                newNums[i] = Math.max(nums[2 * i], nums[2 * i + 1]);
            }
        }
        return minMaxGame(newNums);
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).