Sort Even and Odd Indices Independently

You are given a 0-indexed integer array nums. Rearrange the values of nums according to the following rules:

1. Sort the values at odd indices of nums in non-increasing order.
   • For example, if nums = [4,**1**,2,**3**] before this step, it becomes [4,**3**,2,**1**] after. The values at odd indices 1 and 3 are sorted in non-increasing order.
2. Sort the values at even indices of nums in non-decreasing order.
   • For example, if nums = [**4**,1,**2**,3] before this step, it becomes [**2**,1,**4**,3] after. The values at even indices 0 and 2 are sorted in non-decreasing order.

Return the array formed after rearranging the values of nums.

Example 1:

Input: nums = [4,1,2,3]
Output: [2,3,4,1]
Explanation:
First, we sort the values present at odd indices (1 and 3) in non-increasing order.
So, nums changes from [4,1,2,3] to [4,3,2,1].
Next, we sort the values present at even indices (0 and 2) in non-decreasing order.
So, nums changes from [4,1,2,3] to [2,3,4,1].
Thus, the array formed after rearranging the values is [2,3,4,1].

Example 2:

Input: nums = [2,1]
Output: [2,1]
Explanation:
Since there is exactly one odd index and one even index, no rearrangement of values takes place.
The resultant array formed is [2,1], which is the same as the initial array.

Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100

SOLUTION:

import heapq

class Solution:
    def sortEvenOdd(self, nums: List[int]) -> List[int]:
        n = len(nums)
        evens = []
        odds = []
        for i in range(n):
            if i & 1:
                heapq.heappush(odds, -nums[i])
            else:
                heapq.heappush(evens, nums[i])
        for i in range(n):
            if i & 1:
                nums[i] = -heapq.heappop(odds)
            else:
                nums[i] = heapq.heappop(evens)
        return nums