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
:
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
.i
where 0 <= i < n / 2
, assign the value of newNums[i]
as min(nums[2 * i], nums[2 * i + 1])
.i
where 0 <= i < n / 2
, assign the value of newNums[i]
as max(nums[2 * i], nums[2 * i + 1])
.nums
with newNums
.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 and Description |
---|
Solution() |
Modifier and Type | Method and Description |
---|---|
int |
minMaxGame(int[] nums) |
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