leetcode2293.Min Max Game

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: 1Explanation: 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: 3Explanation: 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.

Example 1:

Java

Runtime24 ms

Beats

5.89%

Memory44.6 MB

Beats

11.44%

Click to check the distribution chart

Notes