Class Solution

java.lang.Object
g2201_2300.s2293_min_max_game.Solution

public class Solution extends Object
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`.
  • Constructor Details

    • Solution

      public Solution()
  • Method Details

    • minMaxGame

      public int minMaxGame(int[] nums)