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 of?nums. If?n == 1,?end?the process. Otherwise,?create?a new?0-indexed?integer array?newNums?of length?n / 2.

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]).

3. 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]).

4. Replace?the array?nums?with?newNums.

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: 1Explanation: 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: 3Explanation: 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.

Example 1:

Java

Runtime24 ms

Beats

5.89%

Memory44.6 MB

Beats

11.44%

Click to check the distribution chart

Notes