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 ofnums
. Ifn == 1
, end the process. Otherwise, create a new 0-indexed integer arraynewNums
of lengthn / 2
. - For every even index
i
where0 <= i < n / 2
, assign the value ofnewNums[i]
asmin(nums[2 * i], nums[2 * i + 1])
. - For every odd index
i
where0 <= i < n / 2
, assign the value ofnewNums[i]
asmax(nums[2 * i], nums[2 * i + 1])
. - Replace the array
nums
withnewNums
. - 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
.
SOLUTION:
class Solution:
def minMax(self, nums, i, j, minm):
if j == i + 1:
return nums[i]
mid = (i + j) // 2
a = self.minMax(nums, i, mid, True)
b = self.minMax(nums, mid, j, False)
if minm:
return min(a, b)
return max(a, b)
def minMaxGame(self, nums: List[int]) -> int:
n = len(nums)
return self.minMax(nums, 0, n, True)
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