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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2201_2300.s2293_min_max_game;

// #Easy #Array #Simulation #2022_06_14_Time_1_ms_(90.39%)_Space_44_MB_(50.37%)

/**
 * 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`:
 *
 * 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._
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2022/04/13/example1drawio-1.png)
 *
 * **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`.
**/
public 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);
    }
}