Package g2201_2300.s2293_min_max_game

Class 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:

    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 Detail

      • Solution

        public Solution()

    • Method Detail

      • minMaxGame

        public int minMaxGame​(int[] nums)