2164. Sort Even and Odd Indices Independently

Description

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

sort-even-and-odd-indices-independently.py

class Solution:
    def sortEvenOdd(self, nums: List[int]) -> List[int]:
        n = len(nums)
        odd, even = [], []

        for i, x in enumerate(nums):
            if i & 1:
                odd.append(x)
            else:
                even.append(x)

        odd.sort(key = lambda x: -x)
        even.sort()

        res = [-1] * n
        index = 0

        for i in range(0, n, 2):
            res[i] = even[index]
            index += 1

        index = 0

        for i in range(1, n, 2):
            res[i] = odd[index]
            index += 1

        return res