s2293_min_max_game.go

/*
	https://leetcode.com/problems/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:

		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.
*/

package solutions

func minMaxGame(nums []int) int {
	for len(nums) > 1 {
		newNums := make([]int, len(nums)/2)
		for i := 0; i < len(newNums); i++ {
			if i%2 == 0 {
				newNums[i] = Minimum(nums[2*i], nums[2*i+1])
			} else {
				newNums[i] = Maximum(nums[2*i], nums[2*i+1])
			}
		}
		nums = newNums
	}
	return nums[0]
}