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 ofnums
. Ifn == 1
, end the process. Otherwise, create a new 0-indexed integer arraynewNums
of lengthn / 2
. - 2.For every even index
i
where0 <= i < n / 2
, assign the value ofnewNums[i]
asmin(nums[2 * i], nums[2 * i + 1])
. - 3.For every odd index
i
where0 <= i < n / 2
, assign the value ofnewNums[i]
asmax(nums[2 * i], nums[2 * i + 1])
. - 4.Replace the array
nums
withnewNums
. - 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 of2
.
func minMaxGame(nums []int) int {
if len(nums) == 1 { return nums[0] }
res := make([]int ,0)
for i:=0; i<len(nums)/2; i++ {
index := i * 2
if i % 2 == 0 {
res = append(res, min(nums[index], nums[index+1]))
} else {
res = append(res, max(nums[index], nums[index+1]))
}
}
return minMaxGame(res)
}
func min(a , b int) int {
if a < b { return a }
return b
}
func max(a , b int) int {
if a > b { return a }
return b
}