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
nbe the length ofnums. Ifn == 1, end the process. Otherwise, create a new 0-indexed integer arraynewNumsof lengthn / 2. - 2.For every even index
iwhere0 <= i < n / 2, assign the value ofnewNums[i]asmin(nums[2 * i], nums[2 * i + 1]). - 3.For every odd index
iwhere0 <= i < n / 2, assign the value ofnewNums[i]asmax(nums[2 * i], nums[2 * i + 1]). - 4.Replace the array
numswithnewNums. - 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 <= 10241 <= nums[i] <= 109nums.lengthis 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
}